POTENTIAL THEORY INSEVERAL COMPLEX VARIABLES

Course of Jean-Pierre DEMAILLYat the ICPAM Summer Schoolon Complex Analysis,Nice, France, July 3–7, 1989

1. Monge-Ampere operators.

Let X be a complex manifold of dimension n . We denote as usual d = d′ + d′′

the exterior derivative and we set

dc =1

2iπ(d′ − d′′) ,

so that ddc = iπd

′d′′ . In this context, we have the following integration by partsformula.

Formula 1.1. — Let Ω ⊂⊂ X be a smooth open subset of X and f, g formsof class C2 on Ω of pure bidegrees (p, p) and (q, q) with p+ q = n− 1 . Then

∫

Ω

f ∧ ddcg − ddcf ∧ g =

∫

∂Ω

f ∧ dcg − dcf ∧ g .

Proof. — By Stokes’ theorem the right hand side is the integral over Ω of

d(f ∧ dcg − dcf ∧ g) = f ∧ ddcg − ddcf ∧ g + (df ∧ dcg + dcf ∧ dg) .

As all forms of total degree 2n and bidegree 6= (n, n) are zero, we get

df ∧ dcg =i

2π(d′f ∧ d′′g − d′′f ∧ d′g) = −dcf ∧ dg .

Let u be a psh function on X and T a closed positive current of bidimension(p, p), i.e. of bidegree (n − p, n − p) . Our desire is to define the wedge productddcu ∧ T even when neither u nor T are smooth. A priori, this product makesno sense since ddcu and T have measure coefficients in general. Assume that uis a locally bounded psh function. Then the current uT is well defined since uis a locally bounded Borel function and T has measure coefficients. According toBedford-Taylor [B-T2] one defines

ddcu ∧ T = ddc(uT )

where ddc( ) is taken in the sense of distribution (or current) theory.

Proposition 1.2. — The wedge product ddcu ∧ T is again a closed positivecurrent.

Proof. — The result is local. We may assume that X is an open set Ω ⊂ Cn ,and after shrinking Ω , that |u| 6 M on Ω . Let (ρε) be a family of regularizingkernels with Supp ρε ⊂ B(0, ε) and

∫ρε = 1 . The sequence of convolutions

uk = u ⋆ ρ1/k is decreasing, bounded by M , and converges pointwise to u ask → +∞ . By Lebesgue’s dominated convergence theorem ukT converges weaklyto uT , thus ddc(ukT ) converges weakly to ddc(uT ) . However, since uk is smooth,ddc(ukT ) coincides with the product ddcuk ∧ T in its usual sense. As T > 0 andas ddcuk is a positive (1, 1)–form, we have ddcuk ∧ T > 0 , hence the weak limitddcu ∧ T is > 0 (and obviously closed).

Given locally bounded psh functions u1, . . . , uq , one defines inductively

ddcu1 ∧ ddcu2 ∧ . . . ∧ dd

cuq ∧ T = ddc(u1ddcu2 . . . ∧ dd

cuq ∧ T )

and the result is a closed positive current. In particular, when u is a locally boundedpsh function, there is a well defined positive measure (ddcu)n . If u is of class C2,a computation in local coordinates gives

(ddcu)n = det( ∂2u

∂zj∂zk

)·n!

πnidz1 ∧ dz1 ∧ . . . ∧ idzn ∧ dzn .

The expression “Monge-Ampere operator” refers generally to the non-linear partialdifferential operator u 7−→ det(∂2u/∂zj∂zk) .

Now, let Θ be a current of order 0 . If K is a compact subset contained in acoordinate patch of X , we define the mass of Θ =

∑ΘI,J dzI ∧ dzJ on K by

||Θ||K =

∫

K

∑

I,J

|ΘI,J |

where |ΘI,J | is the absolute value of the measure ΘI,J . When Θ > 0 we have|ΘI,J | 6 C.Θ ∧ βp with β = ddc|z|2 ; up to constants, the mass ||Θ||K is thenequivalent to the integral

∫K

Θ ∧ βp . When K ⊂⊂ X is arbitrary, we take apartition K =

⋃Kj where each Kj is contained in a coordinate patch and write

||Θ||K =∑

||Θ||Kj.

Up to constants, the semi-norm ||Θ||K does not depend on the choice of thecoordinate systems involved.

Chern-Levine-Nirenberg inequalities 1.3 ([C-L-N]). — For all com-pact sets K,L of X with L ⊂ K , there exists a constant CK,L > 0 such that

||ddcu1 ∧ . . . ∧ ddcuq ∧ T ||L 6 CK,L ||u1||L∞(K) . . . ||uq||L∞(K) ||T ||K .

Proof. — By induction, it is sufficient to prove the result for q = 1 and u1 = u .There is a covering of L by a family of balls B′

j ⊂⊂ Bj ⊂ K contained in coordinatepatches of X . Let χ ∈ D(Bj) be equal to 1 on B

′

j . Then

||ddcu ∧ T ||L∩B

′

j6 C

∫

B′

j

ddcu ∧ T ∧ βp−1 6 C

∫

Bj

χddcu ∧ T ∧ βp−1 .

As T and β are closed, an integration by parts yields

||ddcu ∧ T ||L∩B

′

j6 C

∫

Bj

uT ∧ ddcχ ∧ βp−1 6 C′||u||L∞(K)||T ||K

where C′ is equal to C multiplied by a bound for the coefficients of ddcχ∧βp−1 .

Exercise 1.4. — Denote by L1(K) the space of integrable functions withrespect to some smooth positive density on K . For any V psh on X show

(a) ||ddcV ||L 6 CK,L ||V ||L1(K) .

(b) supLV+ 6 CK,L ||V ||L1(K) .

Now, we prove a rather important continuity theorem due to [B-T2].

Theorem 1.5. — Let u1, . . . , uq be locally bounded psh functions and letuk1 , . . . , u

kq be decreasing sequences of psh functions converging pointwise to

u1, . . . , uq . Then

(a) uk1ddcuk2 ∧ . . . ∧ ddcukq ∧ T −→ u1dd

cu2 ∧ . . . ∧ ddcuq ∧ T weakly.

(b) ddcuk1 ∧ . . . ∧ ddcukq ∧ T −→ ddcu1 ∧ . . . ∧ ddcuq ∧ T weakly.

Proof. — As the sequence (ukj ) is non increasing and as uj is locally bounded,the family (ukj )k∈N is locally uniformly bounded. The result is local, so we can workon a strongly pseudoconvex open set Ω ⊂⊂ X . Let ψ be a strongly psh functionof class C∞ near Ω with ψ < 0 on Ω , ψ = 0 and dψ 6= 0 on ∂Ω . After addition ofa constant we can assume that −M 6 ukj 6 −1 near Ω . Let us denote by (uk,εj ) ,ε ∈]0, ε0] , an increasing family of regularizations converging to ukj as ε → 0 andsuch that −M 6 uk,εj 6 −1 on Ω . Set A = M/δ with δ > 0 small and replace ukjby vkj = maxAψ, ukj , uk,εj by vk,εj = maxεAψ, u

k,εj where maxε = max ⋆ ρε is

a regularized max function.

0

−1

−M

R

Aψ

Ωδ Ω r Ωδ

ukj

Fig. 1 Construction of vkj

Then vkj coincides with ukj on Ωδ = ψ < −δ since Aψ < −Aδ = −M on Ωδ ,and vkj is equal to Aψ on the corona Ω \Ωδ/M . Without loss of generality, we cantherefore assume that all ukj (and similarly all uk,εj ) coincide with Aψ on a fixedneighborhood of ∂Ω .

Now, we argue by induction on q and observe that (b) is an immediateconsequence of (a). When q = 1 , (a) follows directly from the bounded convergencetheorem. We need a lemma.

Lemma 1.6. — Let fk be a non-increasing sequence of upper semi-continuousfunctions converging to f on some separable locally compact space X and µk asequence of positive measures converging weakly to µ on X . Then every weaklimit ν of fkµk satisfies ν 6 fµ .

Indeed if (gp) is a decreasing sequence of continuous functions converging tofk0 for some k0 , then fkµk 6 fk0µk 6 gpµk for k > k0 , thus ν 6 gpµ as k → +∞ .The monotone convergence theorem then gives ν 6 fk0µ as p → +∞ and ν 6 fµas k0 → +∞ .

End of proof of theorem 1.5. — Assume that (a) has been proved for q − 1 .Then

Sk = ddcuk2 ∧ . . . ∧ ddcukq ∧ T −→ S = ddcu2 ∧ . . . ∧ ddcuq ∧ T .

By 1.3 the sequence (uk1Sk) has locally bounded mass, hence is relatively compact

for the weak topology. In order to prove (a), we only have to show that everyweak limit Θ of uk1S

k is equal to u1S . Let (m,m) be the bidimension of S andlet γ be an arbitrary smooth and strongly positive form of bidegree (m,m) . Thenthe positive measures Sk ∧ γ converge weakly to S ∧ γ and lemma 1.6 shows thatΘ ∧ γ 6 u1S ∧ γ , hence Θ 6 u1S . To get the equality, we set β = ddcψ > 0 andshow that

∫Ωu1S ∧ βm 6

∫Ω

Θ ∧ βm , i.e.∫

Ω

u1ddcu2 ∧ . . . ∧ dd

cuq ∧ T ∧ βm 6 lim inf

∫

Ω

uk1ddcuk2 ∧ . . . ∧ ddcukq ∧ T ∧ βm .

As u1 6 uk1 6 uk,ε11 for every ε1 > 0 we get∫

Ω

u1ddcu2 ∧ . . . ∧ dd

cuq ∧ T ∧ βm

6

∫

Ω

uk,ε11 ddcu2 ∧ . . . ∧ ddcuq ∧ T ∧ βm

=

∫

Ω

ddcuk,ε11 ∧ u2ddcu3 ∧ . . . ∧ dd

cuq ∧ T ∧ βm

after an integration by parts (there is no boundary term because uk,ε11 and u2

vanish on ∂Ω). Repeating this argument with u2, . . . , uq , we obtain∫

Ω

u1ddcu2 ∧ . . . ∧ dd

cuq ∧ T ∧ βm

6

∫

Ω

ddcuk,ε11 ∧ . . . ∧ ddcuk,εq−1

q−1 ∧ uqT ∧ βm

6

∫

Ω

uk,ε11 ddcuk,ε22 ∧ . . . ∧ ddcuk,εqq ∧ T ∧ βm .

Now let εq → 0, . . . , ε1 → 0 in this order. We have weak convergence at each stepand uk,ε11 = 0 on the boundary; therefore the last integral converges and we getthe desired inequality

∫

Ω

u1ddcu2 ∧ . . . ∧ dd

cuq ∧ T ∧ βm 6

∫

Ω

uk1ddcuk2 ∧ . . . ∧ ddcukq ∧ T ∧ βm .

Corollary 1.7. — The product ddcu1 ∧ . . . ∧ ddcuq ∧ T is symmetric with

respect to u1, . . . , uq .

Observe that the definition was unsymmetric. The result is true when u1, . . . , uqare smooth and follows in general from theorem 1.5 applied to uk1 = u1 ⋆ ρ1/k .

Theorem 1.8. — Let K,L be compact subsets of X such that L ⊂ K .For any psh functions V, u1, . . . , uq on X such that u1, . . . , uq are locally bounded,there is an inequality

||V ddcu1 ∧ . . . ∧ ddcuq||L 6 CK,L ||V ||L1(K)||u1||L∞(K) . . . ||uq||L∞(K) .

Proof. — First, we may assume that L is contained in a strictly pseudoconvexopen set Ω = ψ < 0 ⊂ K (otherwise cover L by small balls contained in K). Asuitable normalization gives −2 6 uj 6 −1 on K ; then we can modify uj on Ω \Lso that uj = Aψ on Ω \ Ωδ with a fixed constant A and δ > 0 such that L ⊂ Ωδ .Let χ > 0 be a smooth function equal to −ψ on Ωδ with compact support in Ω .If we take ||V ||L1(K) = 1 , we see that V+ is uniformly bounded on Ωδ by 1.4 (b);after subtraction of a fixed constant we get V 6 0 on Ωδ . As uj = Aψ on Ω \Ωδ ,we find for q 6 n− 1 :

∫

Ωδ

− V ddcu1 ∧ . . . ∧ ddcuq ∧ β

n−q

=

∫

Ω

V ddcu1 ∧ . . . ∧ ddcuq ∧ β

n−q−1 ∧ ddcχ−Aq∫

Ω\Ωδ

V βn−1 ∧ ddcχ

=

∫

Ω

χddcV ∧ ddcu1 ∧ . . . ∧ ddcuq ∧ β

n−q−1 −Aq∫

Ω\Ωδ

V βn−1 ∧ ddcχ .

The first integral of the last line is uniformly bounded thanks to 1.3 and 1.4 (a),and the second one is bounded by ||V ||L1(Ω) 6 constant. Inequality 1.8 follows ifq 6 n− 1 . If q = n , we can work instead on X × C and consider V, u1, . . . , uq asfunctions on X × C independent of the extra factor C .

Now, we would like to define ddcu1 ∧ . . . ∧ ddcuq ∧ T also in some cases when

u1, . . . , uq are not bounded below everywhere. Consider first the case q = 1 andlet u be a plurisubharmonic function on X . The polar set of u is by definitionu−1(−∞) .

Assumptions 1.9. — We make two additional assumptions :

(a) T has non zero bidimension (p, p) (i.e. degree of T < 2n) .

(b) X is covered by a family of strongly pseudoconvex open sets Ω = ψ < 0 ,Ω ⊂⊂ X , with the following property : there is an open set ωT containingSupp T ∩ Ω and an open set ωu containing u−1(−∞) ∩ Ω such that ωT ∩ ωu iscompact in Ω and u is bounded on ωT \ ωu .

Example. — For any T , hypothesis 1.9 (b) is clearly satisfied if u has a discreteset P of poles; an interesting example is u = log |F | where F = (F1, . . . , FN ) areholomorphic functions having a discrete set of common zeroes.

Let us replace u by the everywhere finite function

u>s(z) = maxu(z), s .

We shall let hereafter s tend to −∞ . Let β = ddcψ and let s0 be a lower bound foru on a neighborhood of ∂Ω∩Supp T . For s < s0 , the integral

∫Ωddcu>s∧T ∧βp−1

does not depend on s ; in fact, Stokes’ theorem shows that∫

Ω

(ddcu>r − ddcu>s) ∧ T ∧ (ddcψ)p−1 =

∫

Ω

ddc[(u>r − u>s)T ∧ (ddcψ)p−1

]= 0

because u>r and u>s both coincide with u near ∂Ω∩Supp T , hence the current [. . .]has compact support in Ω . This shows that the mass of ddcu>s ∧ T is uniformlybounded on Ω . Now let χ be a function with compact support in Ω equal to ψ ona neighborhood Ω′ of ωT ∩ ωu . As u is bounded on (Ω \ Ω′) ∩ Supp T , we have∫

Ω

χddcu>s∧T ∧(ddcψ)p−1 =

∫

Ω

u>sT∧(ddcψ)p−1∧ddcχ 6 C+

∫

Ω′

u>sT ∧(ddcψ)p.

The first integral remains bounded as s → −∞ . Hence the last integral cannotdecrease to −∞ and we see that uT has bounded mass on Ω′ . We can thereforedefine ddcu ∧ T = ddc(uT ) as before.

Remark 1.10. — The current uT has not necessarily a finite mass when T hasdegree 2n (i.e. T is a measure); example : T = δ0 and u(z) = log |z| in Cn .

Assume now that u1, . . . , uq are psh functions on X that are bounded onωT \ ωu , where ωu is an open set containing all polar sets u−1

j (−∞) such thatωT ∩ ωu ⊂⊂ Ω . One can again use induction to define

(1.11) ddcu1 ∧ ddcu2 ∧ . . . ∧ dd

cuq ∧ T = ddc(u1ddcu2 . . . ∧ dd

cuq ∧ T ) .

Theorem 1.12. — If uk1 , . . . , ukq are non-increasing sequences converging

pointwise to u1, . . . , uq , then

uk1ddcuk2 ∧ . . . ∧ ddcukq ∧ T −→ u1dd

cu2 ∧ . . . ∧ ddcuq ∧ T weakly,

ddcuk1 ∧ . . . ∧ ddcukq ∧ T −→ ddcu1 ∧ . . . ∧ ddcuq ∧ T weakly.

Proof. — Same proof as in theorem 1.5, with the following minor modification :the max procedure maxukj , Aψ is applied only on ωT \Ωδ and ukj is left unchanged

on ωT ∩Ωδ , assuming that Ωδ ⊃ ωT ∩ωu ; observe that the functions ukj and uk,εjare needed only on ωT .

Theorem 1.13. — Let P be a compact subset of a strongly pseudoconvexopen set Ω ⊂ X . If V is a psh function on X and u1, . . . , uq , 1 6 q 6 n− 1 , arepsh functions that are locally bounded on Ω \ P , then V ddcu1 ∧ . . . ∧ ddcuq hasfinite mass on Ω .

Proof. — Same proof as 1.8, taking P ⊂ Ωδ .

2. Generalized Lelong numbers.

Assume from now on that X is a Stein manifold, i.e. that X has a strictly pshexhaustion function. Let ϕ : X −→ [−∞,+∞[ be a continuous psh function. Thesets

S(r) = x ∈ X ; ϕ(x) = r ,(2.1)

B(r) = x ∈ X ; ϕ(x) < r ,(2.1′)

B(r) = x ∈ X ; ϕ(x) 6 r(2.1′′)

will be called pseudo-spheres and pseudo-balls associated to ϕ . It may happenin some cases that B(r) is distinct from the closure of B(r) . For simplicity, wesometimes denote

(2.2) α = ddcϕ , β =1

2ddc(e2ϕ) .

The most simple example we have in mind is ϕ(z) = log |z− a| on an open subsetX ⊂ Cn ; in this case B(r) is the euclidean ball of center a and radius er , and βis the usual hermitian metric i

2πd′d′′|z|2 of Cn . When a = 0 , α is the pull back

on Cn of the standard Fubini-Study metric on Pn−1 .

Definition 2.3. — We say that ϕ is semi-exhaustive if there exists a realnumber R such that B(R) ⊂⊂ X . Similarly, ϕ is said to be semi-exhaustive on aclosed subset A ⊂ X if there exists R such that A ∩B(R) ⊂⊂ X .

We are interested especially in the set of poles S(−∞) = ϕ = −∞ andin the behaviour of ϕ near S(−∞) . Let T be a closed positive current ofbidimension (p, p) on X . Assume that ϕ is semi-exhaustive on Supp T and thatB(R) ∩ Supp T ⊂⊂ X . Then P = S(−∞) ∩ SuppT is compact and the results of§1 show that the measure T ∧ (ddcϕ)p is well defined.

Definition 2.4. — For r ∈] −∞, R[ , we set

ν(T, ϕ, r) =

∫

B(r)

T ∧ (ddcϕ)p ,

ν(T, ϕ) =

∫

S(−∞)

T ∧ (ddcϕ)p = limr→−∞

ν(T, ϕ, r) .

The number ν(T, ϕ) will be called the (generalized) Lelong number of T withrespect to the weight ϕ .

It is clear that r 7−→ ν(T, ϕ, r) is an increasing function of r . Before giving anexample, we need a formula.

Formula 2.5. — For any convex increasing function χ : R −→ R one has∫

B(r)

T ∧ (ddcχ ϕ)p = χ′(r − 0)p ν(T, ϕ, r)

where χ′(r − 0) denotes the left derivative of χ at r .

Proof. — Let χε be the convex function equal to χ on [r−ε,+∞[ and to a linearfunction of slope χ′(r−ε−0) on ]−∞, r−ε] . We get ddc(χεϕ) = χ′(r−ε−0)ddcϕon B(r − ε) and Stokes’ theorem implies∫

B(r)

T ∧ (ddcχ ϕ)p =

∫

B(r)

T ∧ (ddcχε ϕ)p

>

∫

B(r−ε)

T ∧ (ddcχε ϕ)p = χ′(r − ε− 0)pν(T, ϕ, r − ε) .

Similarly, taking χε equal to χ on ]−∞, r− ε] and linear on [r− ε, r] , we obtain∫

B(r−ε)

T ∧ (ddcχ ϕ)p 6

∫

B(r)

T ∧ (ddcχε ϕ)p = χ′(r − ε− 0)pν(T, ϕ, r) .

The expected formula follows when ε tends to 0 .

We get in particular∫B(r)

T∧(ddce2ϕ)p = (2e2r)pν(T, ϕ, r) , whence the formula

(2.6) ν(T, ϕ, r) = e−2pr

∫

B(r)

T ∧ βp .

Now, assume that X is an open subset of Cn and that ϕ(z) = log |z − a| forsome a ∈ X . Formula (2.6) gives

ν(T, ϕ, log r) = r−2p

∫

|z−a|<r

T ∧( i

2πd′d′′|z|2

)p.

The positive measure σT = 1p!T ∧ ( i2d

′d′′|z|2)p = 2−p∑TI,I . i

ndz1 ∧ dz1 . . . dzn iscalled the trace measure of T . We get

(2.7) ν(T, ϕ, log r) =σT

(B(a, r)

)

πpr2p/p!

and ν(T, ϕ) is the limit of this ratio as r → 0 . This limit is called the Lelongnumber of T at point a and denoted ν(T, a) . This was precisely the originaldefinition of Lelong (cf. [Le3]). Let us mention an important consequence.

Consequence 2.8. — The ratio σT(B(a, r)

)/r2p is an increasing function of

the radius r . In particular, we have

σT(B(a, r)

)6 Cr2p

for r < r0 small enough.

All these results are particularly interesting when T = [A] is the current ofintegration over an analytic subset A ⊂ X of pure dimension p . Then σT

(B(a, r)

)

is the euclidean area of A ∩B(a, r) , and ν(T, ϕ, log r) is the ratio of this area tothe area of a ball of radius r in Cp .

Exercise 2.9. — When A is a smooth submanifold of X , show that

ν([A], x) =

1 for x ∈ A0 for x /∈ A .

Remark 2.10. — When X = Cn , ϕ(z) = log |z − a| and A = X (i.e. T = 1),we obtain in particular

∫B(a,r)

(ddc log |z − a|)n = 1 for all r . This implies

(ddc log |z − a|)n = δa .

This fundamental formula can be viewed as a higher dimensional analogue of theusual formula ∆ log |z − a| = 2πδa in C .

3. The Lelong-Jensen formula.

Assume in this paragraph that ϕ is semi-exhaustive onX and that B(R) ⊂⊂ X .For every r ∈] − ∞, R[ , the measures ddc(ϕ>r)

n are well defined. The mapr 7−→ (ddcϕ>r)

n is continuous on ] − ∞, R[ with respect to the weak topology :right continuity follows immediately from theorem 1.5, while left continuity isobtained similarly from the equality (ddcϕ>r)

n = (ddcmaxϕ − r, 0)n . As(ddcϕ>r)

n = (ddcϕ)n on X \ B(r) and ϕ>r ≡ r , (ddcϕ>r)n = 0 on B(r) ,

the left continuity implies (ddcϕ>r)n > 1X\B(r)(dd

cϕ)n . Here 1A denotes thecharacteristic function of any subset A ⊂ X . According to the definitionintroduced in [De2], the collection of Monge-Ampere measures associated to ϕis the family of positive measures µr such that

(3.1) µr = (ddcϕ>r)n − 1X\B(r)(dd

cϕ)n , r ∈] −∞, R[ .

The measure µr is supported on S(r) and r 7−→ µr is weakly continuouson the left by the bounded convergence theorem. Stokes’ formula shows that∫B(s)

(ddcϕ>r)n−(ddcϕ)n = 0 for s > r , hence the total mass µr

(S(r)

)= µr

(B(s)

)

is equal to the difference between the masses of (ddcϕ)n and 1X\B(r)(ddcϕ)n over

B(s) , i.e.

(3.2) µr(S(r)

)=

∫

B(r)

(ddcϕ)n .

Example 3.3. — When (ddcϕ)n = 0 on X \ϕ−1(−∞), formula (3.1) simplifiesinto µr = (ddcϕ>r)

n . This is so for ϕ(z) = log |z| . In this case, the invarianceof ϕ under unitary transformations implies that µr is also invariant. As the totalmass of µr is equal to 1 by 2.10 and (3.2), we see that µr is the invariant measureof mass 1 on the euclidean sphere of radius er .

Theorem 3.4. — Assume that ϕ is smooth near S(r) and that dϕ 6= 0 onS(r) , i.e. r is a non critical value. Then S(r) = ∂B(r) is a smooth oriented realhypersurface and µr is given by the (2n− 1)–volume form (ddcϕ)n−1 ∧ dcϕS(r) .

Proof. — Write maxt, r = limk→+∞ χk(t) where χ is a non-increasingsequence of smooth convex functions with χk(t) = r for t 6 r− 1/k , χk(t) = t fort > r+1/k . Theorem 1.5 shows that (ddcχk ϕ)n converges weakly to (ddcϕ>r)

n .Let h be a smooth function h with compact support near S(r) . Let us applyStokes’ theorem with S(r) considered as the boundary of X \B(r) :

∫

X

h(ddcϕ>r)n = lim

k→+∞

∫

X

h(ddcχk ϕ)n

= limk→+∞

∫

X

−dh ∧ (ddcχk ϕ)n−1 ∧ dc(χk ϕ)

= limk→+∞

∫

X

−χ′k(t)

n dh ∧ (ddcϕ)n−1 ∧ dcϕ

=

∫

X\B(r)

−dh ∧ (ddcϕ)n−1 ∧ dcϕ

=

∫

S(r)

h (ddcϕ)n−1 ∧ dcϕ+

∫

X\B(r)

h (ddcϕ)n−1 ∧ dcϕ .

Near S(r) we thus have an equality of measures

(ddcϕ>r)n = (ddcϕ)n−1 ∧ dcϕS(r) + 1X\B(r)(dd

cϕ)n .

Lelong-Jensen formula 3.5. — Let V be any psh function on X . ThenV is µr–integrable for every r ∈] −∞, R[ and

µr(V ) −

∫

B(r)

V (ddcϕ)n =

∫ r

−∞

ν(ddcV, ϕ, t) dt .

Proof. — Theorem 1.8 shows that V is integrable with respect to (ddcϕ>r)n ,

hence V is µr–integrable. By definition ν(ddcV, ϕ, t) =∫ϕ(z)<t

ddcV ∧ (ddcϕ)n−1

and Fubini’s theorem gives∫ r

−∞

ν(ddcV, ϕ, t) dt =

∫∫

ϕ(z)<t<r

ddcV (z) ∧ (ddcϕ(z))n−1 dt

=

∫

B(r)

(r − ϕ)ddcV ∧ (ddcϕ)n−1 .(3.6)

We first show that formula 3.5 is true when ϕ and V are smooth. As both membersof the formula are left continuous with respect to r and as almost all values of ϕare non critical by Sard’s theorem, we may assume r non critical. Formula 1.1applied with f = (r−ϕ)(ddcϕ)n−1 and g = V shows that integral (3.6) is equal to

∫

S(r)

V (ddcϕ)n−1 ∧ dcϕ−

∫

B(r)

V (ddcϕ)n = µr(V ) −

∫

B(r)

V (ddcϕ)n .

Formula 3.5 is thus proved when ϕ and V are smooth. If V is smooth and ϕmerely continuous and finite, one can write ϕ = limϕk where ϕk is a non-increasing sequence of smooth plurisubharmonic functions (because X is Stein).

Then ddcV ∧(ddcϕk)n−1 converges weakly to ddcV ∧(ddcϕ)n−1 and (3.6) converges,

since 1B(r)(r − ϕ) is continuous with compact support on X . The left hand sideof formula 3.5 also converges because the definition of µr implies

µk,r(V ) −

∫

ϕk<r

V (ddcϕk)n =

∫

X

V((ddcϕk,>r)

n − (ddcϕk)n)

and we can apply again weak convergence on a neighborhood of B(r) . If ϕ takes−∞ values, replace ϕ by ϕ>−k where k → +∞ . Then µr(V ) is unchanged,∫B(r)

V (ddcϕ>−k)n converges to

∫B(r)

V (ddcϕ)n and the right hand side of formula

3.5 is replaced by∫ r−kν(ddcV, ϕ, t) dt . Finally, for V arbitrary, write V = lim ↓ Vk

with a sequence of smooth functions Vk . Then ddcVk∧(ddcϕ)n−1 converges weaklyto ddcV ∧(ddcϕ)n−1 by theorem 1.13, thus integral (3.6) converges to the expectedlimit, and the same is true for the left hand side of 3.5 by the monotone convergencetheorem.

For r < r0 < R , the Lelong-Jensen formula implies

(3.7) µr(V ) − µr0(V ) +

∫

B(r0)\B(r)

V (ddcϕ)n =

∫ r

r0

ν(ddcV, ϕ, t) dt .

Corollary 3.8. — Assume that (ddcϕ)n = 0 on X \ S(−∞) . Thenr 7−→ µr(V ) is a convex increasing function of r and the lelong number ν(ddcV, ϕ)is given by

ν(ddcV, ϕ) = limr→−∞

µr(V )

r.

Proof. — By (3.7) we have

µr(V ) = µr0(V ) +

∫ r

r0

ν(ddcV, ϕ, t) dt .

As ν(ddcV, ϕ, t) is increasing and non-negative, it follows that r 7−→ µr(V ) isconvex and increasing. The formula for ν(ddcV, ϕ) = limt→−∞ ν(ddcV, ϕ, t) is thenobvious.

Example 3.9. — Take ϕ(z) = log |z − a| on an open subset of Cn containingthe point a . The Lelong-Jensen formula becomes

µr(V ) = V (a) +

∫ r

−∞

ν(ddcV, ϕ, t) dt .

As µr is the mean value measure on the sphere S(a, er) , we make the change ofvariables r 7→ log r , t 7→ log t and obtain the more familiar formula

µ(V, S(a, r)) = V (a) +

∫ r

0

ν(ddcV, t)dt

t

where ν(ddcV, t) = ν(ddcV, ϕ, log t) is given by (2.7) :

ν(ddcV, t) =1

πn−1t2n−2/(n− 1)!

∫

B(a,t)

1

2π∆V .

In particular, take V = log |f | where f is a holomorphic function on X .The Poincare-Lelong formula shows that ddc log |f | is equal to the zero divisor[Zf ] =

∑mj [Hj ] , where Hj are the irreducible components of f−1(0) and mj the

multiplicity of f on Hj . The trace 12π

∆f is then the euclidean area measureof Zf (with corresponding multiplicities mj). In dimension n = 1 , we have12π

∆f =∑mjδaj

. Then we get the usual Jensen formula

µ(log |f |, S(0, r)

)− log |f(0)| =

∫ r

0

ν(t)dt

t=

∑mj log

r

|aj|

where ν(t) is the number of zeroes aj in the disk D(0, t) , counted with multi-plicities mj .

Example 3.10. — Take ϕ(z) = log max |zj |λj . where λj > 0 . Then B(r) is

the polydisk of radii (er/λ1 , . . . , er/λn) . If some coordinate zj is non zero, say z1 ,we can write ϕ(z) as λ1 log |z1| plus some function depending only on the (n− 1)

variables zj/zλ1/λj

1 . Hence (ddcϕ)n = 0 on Cn \ 0 . It will be shown later that

(3.11) (ddcϕ)n = λ1 . . . λn δ0 .

We now determine the measures µr . At any point z where not all terms |zj |λj are

equal, the smallest one can be omitted without changing ϕ in a neighborhood ofz . Thus ϕ depends only on (n− 1)–variables and (ddcϕ>r)

n = 0 , µr = 0 near z .It follows that µr is supported by the distinguished boundary |zj | = er/λj of thepolydisk B(r) . As ϕ is invariant by all rotations zj 7−→ eiθjzj , the measure µr isalso invariant and we see that µr is a constant multiple of dθ1 . . . dθn . By formula(3.2) and (3.11) we get

µr = λ1 . . . λn (2π)−ndθ1 . . . dθn .

In particular, the Lelong number ν(ddcV, ϕ) is given by

ν(ddcV, ϕ) = λ1 . . . λn limr→−∞

∫

θj∈[0,2π]

V (er/λ1+iθ1 , . . . , er/λn+iθn)dθ1 . . . dθn

(2π)n.

These numbers have been introduced and studied by Kiselman [Ki3] ; we shalldenote them ν(T, x, λ) , where λ = (λ1, . . . , λn) .

4. Comparison theorem for Lelong numbers.

We show here that the Lelong numbers ν(T, ϕ) only depend on the asymptoticbehaviour of ϕ near the polar set S(−∞) . In a precise way :

Theorem 4.1. — Let ϕ, ψ : X −→ [−∞,+∞[ be continuous psh functions.We assume that ϕ, ψ are semi-exhaustive on Supp T and that

l := lim supψ(x)

ϕ(x)< +∞ as x ∈ Supp T and ϕ(x) → −∞ .

Then ν(T, ψ) 6 lpν(T, ϕ) , and the equality holds if l = limψ/ϕ .

Proof. — Definition 2.4 gives immediately

ν(T, λϕ) = λpν(T, ϕ)

for every scalar λ > 0 . It is thus sufficient to verify the inequality ν(T, ψ) 6 ν(T, ϕ)under tha hypothesis lim supψ/ϕ < 1 . For all c > 0 , consider the psh function

uc = max(ψ − c, ϕ) .

Let Rϕ and Rψ be such that Bϕ(Rϕ)∩SuppT and Bψ(Rψ)∩SuppT be relativelycompact in X . Let r < Rϕ be fixed and a < r . For c > 0 large enough, we haveuc = ϕ on ϕ−1([a, r]) and Stokes’ formula gives

ν(T, ϕ, r) = ν(T, uc, r) > ν(T, uc) .

The hypothesis lim supψ/ϕ < 1 implies on the other hand that there exists t0 < 0such that uc = ψ − c on uc < t0 ∩ Supp T . We infer

ν(T, uc) = ν(T, ψ − c) = ν(T, ψ) ,

hence ν(T, ψ) 6 ν(T, ϕ) . The equality case is obtained by reversing the roles of ϕand ψ and observing that limϕ/ψ = 1/l .

Assume in particular that zk = (zk1 , . . . , zkn) , k = 1, 2 , are coordinate systems

centered at a point x ∈ X and let

ϕk(z) = log |zk| = log(|zk1 |

2 + . . .+ |zkn|2)1/2

.

We have limz→x ϕ2(z)/ϕ1(z) = 1 , hence ν(T, ϕ1) = ν(T, ϕ2) by theorem 4.1 .

Corollary 4.2. — The usual Lelong numbers ν(T, x) are independent ofthe choice of local coordinates.

This result had been originally proved by [Siu] with a much more delicate proof.Another interesting consequence is :

Corollary 4.3. — On an open subset of Cn , the Lelong numbers andKiselman numbers are related by

ν(T, x) = ν(T, x, (1, . . . , 1)

).

Proof. — By definition, the number ν(T, x) is associated to the weightϕ(z) = log |z − x| and ν

(T, x, (1, . . . , 1)

)to the weight ψ(z) = log max |zj − xj | .

It is clear that limz→x ψ(z)/ϕ(z) = 1 , whence the conclusion.

Another consequence of theorem 4.1 is that ν(T, x, λ) is an increasing functionof each variable λj . Moreover, if λ1 6 . . . 6 λn , we get the inequalities

λp1ν(T, x) 6 ν(T, x, λ) 6 λpnν(T, x) .

It can be shown ([De1]) that the stronger inequalities

λ1 . . . λpν(T, x) 6 ν(T, x, λ) 6 λn−p+1 . . . λnν(T, x)

hold for every current T of bidimension (p, p) . By formula (3.11), this is easilychecked for any p–plane of coordinates T = [Cei1 ⊕ . . .⊕ Ceip ] . The general casecan be deduced from this special case.

Now, we assume that T = [A] is the current of integration over an analytic setA ⊂ X of pure dimension p (cf. P. Lelong[Le1]). The above comparison theoremwill enable us to give a simple proof of P. Thie’s main result [Th] : the Lelongnumber ν([A], x) can be interpreted as the multiplicity of the analytic set A atpoint x .

Let x ∈ A be a given point and IA,x the ideal of germs of holomorphic functionsat x vanishing on A . Then, one can find local coordinates z = (z1, . . . , zn) on Xcentered at x such that there exist distinguished Weierstrass polynomials Pj ∈ IA,xin the variable zj , p < j 6 n , of the type

(4.4) Pj(z) = zdj

j +

dj∑

k=1

aj,k(z1, . . . , zj−1)zdj−kj , aj,k ∈ Mk

Cj−1,0

where MX,x is the maximal ideal of X at x .

Indeed, let us prove this property by induction on codimX = n− p . We fix acoordinate system (w1, . . . , wn) by which we identify the germ (X, x) to (Cn, 0) .

If n− p > 1 , there exists a non zero element f ∈ IA,x . Let d be the smallestinteger such that f ∈ Md

Cn,0 and let en ∈ Cn be a non zero vector such that

limt→0 f(ten)/td 6= 0 . Complete en into a basis (e1, . . . , en−1, en) of Cn and denote

by (z1, . . . , zn−1, zn) the corresponding coordinates. The Weierstrass preparationtheorem gives a factorization f = gP where P is a distinguished polynomial oftype (4.4) in the variable zn and where g is an invertible holomorphic function atpoint x . If n− p = 1 , the polynomial Pn = P satisfies the requirements.

If n−p > 2 , OA,x = OX,x/IA,x is a OCn−1,0 = Cz1, . . . , zn−1-module of finitetype, i.e. the projection pr : (X, x) ≈ (Cn, 0) −→ (Cn−1, 0) is a finite morphism of(A, x) onto a germ (Z, 0) ⊂ (Cn−1, 0) of dimension p . The induction hypothesisapplied to IZ,0 = OCn−1,0∩IA,x implies the existence of a new basis (e1, . . . , en−1)of Cn−1 and polynomials Pp+1, . . . , Pn−1 ∈ IZ,0 of type (4.4) in the coordinates(z1, . . . , zn−1) associated to the basis (e1, . . . , en−1) . If we choose Pn = P , theexpected property is proved in codimension n− p .

For any polynomial Q(w) = wd + a1wd−1 + . . .+ ad ∈ C[w] , the roots w of Q

satisfy

(4.5) |w| 6 2 max16k6d

|ak|1/k,

otherwise Q(w)w−d = 1+a1w−1 + . . .+adw

−d would have a modulus larger than1 − (2−1 + . . . + 2−d) = 2−d , a contradiction. Let us denote z = (z′, z′′) withz′ = (z1, . . . , zp) and z′′ = (zp+1, . . . , zn) . As aj,k ∈ Mk

Cj−1,0, we get

|aj,k(z1, . . . , zj−1)| = O((|z1| + . . .+ |zj−1|)

k)

if j > p ,

and we deduce from (4.4), (4.5) that |zj | = O(|z1| + . . . + |zj−1|) on (A, x) .Therefore, the germ (A, x) is contained in a cone |z′′| 6 C|z′| .

We shall use this property in order to compute the Lelong number of [A] atpoint x . When z ∈ A tends to x , the functions

ϕ(z) = log |z| = log(|z′|2 + |z′′|2)1/2 , ψ(z) = log |z′| .

are equivalent. As ϕ, ψ are semi-exhaustive on A , theorem 4.1 implies

ν([A], x) = ν([A], ϕ) = ν([A], ψ).

Let B′ ⊂ Cp the ball of center 0 and radius r′ , B′′ ⊂ Cn−p the ball of center 0and radius r′′ = Cr′ . The inclusion of germ (A, x) in the cone |z′′| 6 C|z′| showsthat for r′ small enough the projection

pr : A ∩ (B′ ×B′′) −→ B′

is proper. The fibers are finite by (4.4). Hence this projection is a ramified coveringwith a finite sheet number m .

z′′ ∈ Cn−p

z′ ∈ Cp

A

S S

0

B′

B′′ π

Fig. 2 Ramified covering from A to ∆′ ⊂ Cp.

Let us apply formula (2.6) to ψ : for every t < r′ we get

ν([A], ψ, log t) = t−2p

∫

ψ<log t

[A] ∧(1

2ddce2ψ

)p

= t−2p

∫

A∩|z′|<t

(1

2pr⋆ddc|z′|2

)p

= mt−2p

∫

Cp∩|z′|<t

(1

2ddc|z′|2

)p= m ,

hence ν(T, ψ) = m . Here, we used the fact that pr is actually a covering withm sheets over the complement of the ramification locus S ⊂ B′ , which is of zeroLebesgue measure. We thus obtain :

Theorem 4.6 (P. Thie [Th]). — Let A be an analytic set of dimension p ina complex manifold of dimension p . For every point x ∈ A , there exist localcoordinates

z = (z′, z′′) , z′ = (z1, . . . , zp) , z′′ = (zp+1, . . . , zn)

centered at x and balls B′ ⊂ Cp , B′′ ⊂ Cn−p in of radii r′, r′′ in these coordinates,such that A∩(B′×B′′) is contained in the cone |z′′| 6 (r′′/r′)|z′| . The multiplicityof A at x is defined as the number m of sheets of the ramified covering mapA ∩ (B′ ×B′′) −→ B′ . Then ν([A], x) = m .

Exercise 4.7. — Show that the measures(ddc log max |zj |

λj)n

,(ddc log

∑|zj |

λj

)n

are equal to c(λ)δ0 with the same coefficient c(λ) in both cases. When all λj areeven integers, compute explicitly c(λ) by means of formula (2.6). Extend the resultto arbitrary rational (resp. real) numbers λj > 0 .

5. Siu’s semi-continuity theorem.

Let X, Y be complex manifolds of dimension n, n′ such that X is Stein. Letϕ : X × Y −→ [−∞,+∞[ be a continuous psh function. We assume that ϕ issemi-exhaustive with respect to Supp T , i.e. that for every compact subset L ⊂ Ythere exists R = R(L) < 0 such that

(5.1) (x, y) ∈ Supp T × L ; ϕ(x, y) 6 R ⊂⊂ X × Y.

Let T be a closed positive current of bidimension (p, p) on X . For every pointy ∈ Y , the function ϕy(x) := ϕ(x, y) is semi-exhaustive on Supp T ; one cantherefore associate to y a generalized Lelong number ν(T, ϕy) .

Lemma 5.2. — (a) The measure T ∧ (ddcϕy,>t)p depends continuously on y.

(b) For all r1 < r2 < R(L) and y, y0 ∈ L one has

lim supy→y0

ν(T, ϕy, r1) 6 ν(T, ϕy0 , r2) .

(c) The map y 7→ ν(T, ϕy) is upper semi-continuous.

Proof. — (a) We prove by induction on q that T ∧ (ddcϕy,>t)q is weakly

continuous in y . Let h be a smooth form on X . Then∫

X

h ∧ T ∧ (ddcϕy,>t)q =

∫

X

ddch ∧ T ∧ ϕy,>t(ddcϕy,>t)

q−1 .

Taking the difference for two points y, y0 , we get∫

X

h ∧ T∧((ddcϕy,>t)

q − (ddcϕy0,>t)q)

=

∫

X

ddch ∧ T ∧ (ϕy,>t − ϕy0,>t)(ddcϕy,>t)

q−1)

+

∫

X

ϕy0,>tddch ∧ T ∧

((ddcϕy,>t)

q−1 − (ddcϕy0,>t)q−1

).

The last integral tends to 0 thanks to the induction hypothesis, and the previousone thanks to the Chern-Levine Nirenberg inequalities and the uniform conver-gence of ϕy,>t to ϕy0,>t .

(b) As ν(T, ϕy, r) =∫B(r)

T ∧ (ddcϕy,>t)p when t < r , (b) follows from (a).

(c) Letting r1, r2 → −∞ , we get easily lim supy→y0 ν(T, ϕy) 6 ν(T, ϕy0) .

As a consequence of 5.2 (c), we see that the sublevel sets

(5.3) Ec = y ∈ Y ; ν(T, ϕy) > c , c > 0

are closed. Under mild additional hypotheses, we are going to show that the setsEc are in fact analytic subsets of Y .

Definition 5.3. — We say that a function f(x, y) is locally Holder continu-ous with respect to y on X × Y if every point of X × Y has a neighborhood Ω onwhich

|f(x, y1) − f(x, y2)| 6 M |y1 − y2|γ

for all (x, y1) ∈ Ω , (x, y2) ∈ Ω , for some constants M > 0 , γ ∈]0, 1] , and forsuitable coordinates on Y .

Theorem 5.4. — Let T be a closed positive current on X and

ϕ : X × Y −→ [−∞,+∞[

a continuous psh function. Assume that ϕ is semi-exhaustive on Supp T and thateϕ(x,y) is locally Holder continuous with respect to y on X×Y . Then the sublevelsets

Ec = y ∈ Y ; ν(T, ϕy) > c

are analytic subsets of Y .

This theorem, proved in [De3], can be rephrased by saying that y 7−→ ν(T, ϕy)is upper semi-continuous with respect to the analytic Zariski topology. As a specialcase, we get the following important result of [Siu] :

Corollary 5.5. — If T is a closed positive current on a manifold X , thesublevel sets Ec = x ∈ X ; ν(T, x) > c of the usual Lelong numbers are analytic.

Proof. — The result is local, so we may assume that X ⊂ Cn is an opensubset. Then apply theorem 5.4 with Y = X and ϕ(x, y) = log |x− y| .

IfX is an open subset of Cn , the sublevel sets for Kiselman’s numbers ν(T, x, λ)

are also analytic in X . However, this result is not intrinsically significant on amanifold, because Kiselman’s numbers depend on the choice of coordinates. To giveanother application, set uλ(z) = log max |zj |

λj and set ϕ(x, y, g) = uλ(g(x − y)

)

where x, y ∈ Cn and g ∈ Gl(Cn) . Then ν(T, ϕy,g) is the Kiselman number of Tat y when the coordinates have been rotated by g . It is clear that ϕ is psh in(x, y, g) and semi-exhaustive with respect to x , and that eϕ is Holder continuouswith exponent γ = min1, λj . Thus the sublevel sets

Ec = (y, g) ∈ X × Gl(Cn) ; ν(T, ϕy,g) > c

are analytic in X ×Gl(Cn) . Theorem 5.4 can be applied more generally to weightfunctions of the type

ϕ = maxj

log(∑

k

|Fj,k|λj,k

)

where Fj,k are holomorphic functions on X × Y and γj,k positive real constants;in this case eϕ is Holder continuous of exponent γ = minλj,k, 1 .

Now let us prove theorem 5.4. As the result is local on Y , we may assumewithout loss of generality that Y is a ball in Cn

′

. After addition of a constant toϕ , we may also assume that there exists a compact subset K ⊂ X such that

(x, y) ∈ X × Y ;ϕ(x, y) 6 0 ⊂ K × Y.

By theorem 4.1, the Lelong numbers depend only on the asymptotic behaviour ofϕ near the (compact) polar set ϕ−1(−∞) ∩ (SuppT × Y ) . We can add a smoothstrictly plurisubharmonic function on X × Y to make ϕ strictly plurisuharmonic.Then Richberg’s approximation theorem for continuous psh functions shows thatthere exists a smooth psh function ϕ such that ϕ 6 ϕ 6 ϕ+ 1 . We may thereforeassume that ϕ is smooth on (X × Y ) \ ϕ−1(−∞) .

• First step : construction of a local psh potential.

Our goal is to generalize the usual construction of psh potentials associated toa closed positive current (cf. P. Lelong[Le2] and H. Skoda[Sk]). We replace herethe usual kernel |z − ζ|−2p arising from the hermitian metric of Cn by a kerneldepending on the weight ϕ . Let χ ∈ C∞(R,R) be an increasing function suchthat χ(t) = t for t 6 −1 and χ(t) = 0 for t > 0 . We consider the half-planeH = z ∈ C ; Rez < −1 and associate to T the potential function V on Y ×Hdefined by

(5.6) V (y, z) = −

∫ 0

Rez

ν(T, ϕy, t)χ′(t) dt .

For every t > Re z , Stokes’ formula gives

ν(T, ϕy, t) =

∫

ϕ(x,y)<t

T (x) ∧ (ddcxϕ(x, y, z))p

with ϕ(x, y, z) := maxϕ(x, y),Rez . Fubini’s theorem applied to (5.6) gives

V (y, z) = −

∫

x∈X,ϕ(x,y)<tRe z<t<0

T (x) ∧ (ddcxϕ(x, y, z))p χ′(t)dt

=

∫

x∈X

T (x) ∧ χ(ϕ(x, y, z)) (ddcxϕ(x, y, z))p .

For all (n− 1, n− 1)–form h of class C∞ with compact support in Y ×H , we get

〈ddcV, h〉 = 〈V, ddch〉 =

∫

X×Y×H

T (x) ∧ χ(ϕ(x, y, z))(ddcϕ(x, y, z))p ∧ ddch(y, z).

Observe that the replacement of ddcx by the total differentiation ddc = ddcx,y,z doesnot modify the integrand, because the terms in dx , dx must have total bidegree(n, n) . The current T (x)∧χ(ϕ(x, y, z))h(y, z) has compact support in X×Y ×H.An integration by parts can thus be performed to obtain

〈ddcV, h〉 = (2π)−p∫

X×Y×H

T (x) ∧ ddc(χ ϕ(x, y, z)) ∧ (ddcϕ(x, y, z))p.h(y, z).

On the corona −1 6 ϕ(x, y) 6 0 we have ϕ(x, y, z) = ϕ(x, y) , whereas forϕ(x, y) < −1 we get ϕ < 1 and χ ϕ = ϕ . As ϕ is psh, we see that ddcV (y, z) isthe sum of the positive (1, 1)–form

(y, z) 7−→

∫

x∈X;ϕ(x,y)<−1

T (x) ∧ (ddcx,y,zϕ(x, y, z))p+1

and of the (1, 1)–form independent of z

y 7−→

∫

x∈X;−16ϕ(x,y)60

T ∧ ddcx,y(χ ϕ) ∧ (ddcx,yϕ)p ;

as ϕ is smooth outside ϕ−1(−∞) , this last form has locally bounded coefficients.We obtain therefore the following result.

Theorem 5.7. — There exists a positive psh function ρ ∈ C∞(Y ) such thatρ(y) + V (y, z) is psh on Y ×H .

To be quite complete, we must observe in addition that V is continuous onY × H because T ∧ (ddcϕy,z)

p is weakly continuous in the variables (y, z) bylemma 5.1 (a).

If we let Re z tend to −∞ , we see that the function

U0(y) = ρ(y) + V (y,−∞) = ρ(y) −

∫ 0

−∞

ν(T, ϕy, t)χ′(t)dt

is locally psh or ≡ −∞ on Y . Moreover, it is clear that U0(y) = −∞ at everypoint y such that ν(T, ϕy) > 0 . If Y is connected and U0 6≡ −∞ , we alreadyconclude that the density set

⋃c>0Ec is pluripolar in Y .

• Second step : application of Kiselman’s minimum principle.

Kiselman’s minimum principle 5.8 ([Ki1]). — Let M be a complex man-ifold, ω ⊂ Rn a convex open subset and Ω the “tube domain” Ω = ω + iRn . Forevery plurisubharmonic function v(ζ, z) on M ×Ω that does not depend on Im z ,the function

u(ζ) = infz∈Ω

v(ζ, z)

is plurisubharmonic or locally ≡ −∞ on M .

Proof. — Without loss of generality, we may assume that M is a ball andthat 0 ∈ M × Ω . The hypothesis implies that v(ζ, z) is convex in x = Re z . Letτ > 0 be a smooth strictly convex exhaustion function on ω with τ(0) = 0 . Weapproximate v by the sequence of smooth functions defined on (1 − ε)M × Ω by

vε = v ⋆ ρε2 + ετ((1 − ε)−1x

)

Then uε(ζ) = infz∈Ω vε(ζ, z) is increasing in ε and converges to u . We maytherefore assume that v has all properties of vε , i.e. v is smooth, plurisubharmonicin (ζ, z) , strictly convex in x and lim|x|→+∞ v(ζ, x) = +∞ for every ζ ∈M . Thenx 7−→ v(ζ, x) has a unique minimum point x = g(ζ) , solution of the equations∂v/∂xj(x, ζ) = 0 . As the matrix (∂2v/∂xj∂xk) is positive definite, the implicitfunction theorem shows that g is smooth. Now, if w 7−→ ζ0 +wa , a ∈ Cn , |w| 6 1is a complex disk ∆ contained in M , there exists a holomorphic function f onthe unit disk and smooth up to the boundary, whose real part solves the Dirichletproblem

Re f(eiθ) = g(ζ0 + eiθa) .

Since v(ζ0 + wa, f(w)) is subharmonic in w , we get the mean value inequality

v(ζ0, f(0)) 61

2π

∫ 2π

0

v(ζ0 + eiθa, f(eiθ)

)dθ =

1

2π

∫

∂∆

v(ζ, g(ζ))dθ .

The last equality holds because Re f = g on ∂∆ and v(ζ, z) = v(ζ,Re z) byhypothesis. As u(ζ0) 6 v(ζ0, f(0)) and u(ζ) = v(ζ, g(ζ)) , we see that u satisfiesthe mean value inequality.

Let a > 0 be arbitrary. The function

Y ×H ∋ (y, z) 7−→ ρ(y) + V (y, z) − aRez

is psh and independent of Im z . By 5.8, the Legendre transform

Ua(y) = infr<−1

[ρ(y) + V (y, r)− ar]

is locally psh or ≡ −∞ on Y .

Lemma 5.9. — Let y0 ∈ Y be a given point.

(a) If a > ν(T, ϕy0) , then Ua is bounded below on a neighborhood of y0 .

(b) If a < ν(T, ϕy0) , then Ua(y0) = −∞ .

Proof. — By definition of V (cf. (5.6)) we have

(5.10) V (y, r) 6 −ν(T, ϕy , r)

∫ 0

r

χ′(t)dt = rν(T, ϕy, r) 6 rν(T, ϕy) .

Then clearly Ua(y0) = −∞ if a < ν(T, ϕy0) . On the other hand, if ν(T, ϕy0) < a ,there exists t0 < 0 such that ν(T, ϕy0 , t0) < a . Fix r0 < t0 . The semi-continuity property 5.2 (b) shows that there exists a neighborhood ω of y0 suchthat supy∈ω ν(T, ϕy, r0) < a . For all y ∈ ω , we get

V (y, r) > −C − a

∫ r0

r

χ′(t)dt = −C + a(r − r0) ,

and this implies Ua(y) > −C − ar0 .

Theorem 5.11. — If Y is connected and if Ec 6= Y , then Ec is a closedcomplete pluripolar subset of Y , i.e. there exists a continuous psh functionw : Y −→ [−∞,+∞[ such that Ec = w−1(−∞) .

Proof. — We first observe that the family (Ua) is increasing in a , thatUa = −∞ on Ec for all a < c and that supa<c Ua(y) > −∞ if y ∈ Y \ Ec (applylemma 5.9). For any integer k > 1 , let wk ∈ C∞(Y ) be a psh regularization ofUc−1/k such that wk > Uc−1/k on Y and wk 6 −2k on Ec ∩ Yk . Then lemma5.10 (a) shows that the family (wk) is uniformly bounded below on every compactsubset of Y \ Ec . We can also choose wk uniformly bounded above on everycompact subset of Y because Uc−1/k 6 Uc . The function

w =+∞∑

k=1

2−kwk

satifies our requirements.

• Third step : estimation of the singularities of the potentials Ua.

Lemma 5.12. — Let y0 ∈ Y be a given point, L a compact neighborhood ofy0 , K ⊂ X a compact subset and r0 a real number < −1 such that

(x, y) ∈ X × L;ϕ(x, y) 6 r0 ⊂ K × L .

Assume that eϕ(x, y) is locally Holder continuous in y and that

|f(x, y1) − f(x, y2)| 6 M |y1 − y2|γ

for all (x, y1, y2) ∈ K × L× L . Then, for all ε ∈]0, 1[ , there exists a real numberη(ε) > 0 such that all y ∈ Y with |y − y0| < η(ε) satisfy

Ua(y) 6 ρ(y) +((1 − ε)pν(T, ϕy0) − a

)(γ log |y − y0| + log

2eM

ε

).

Proof. — First, we try to estimate ν(T, ϕy, r) when y ∈ L is near y0 . Set

ψ(x) = (1 − ε)ϕy0(x) + εr − ε/2

ψ(x) = max(ϕy(x), (1 − ε)ϕy0(x) + εr − ε/2

)

ψ(x) = ϕy(x)

if

if

if

ϕy0(x) 6 r − 1

r − 1 6ϕy0(x) 6 r

r 6ϕy0(x) 6 r0

and verify that this definition is coherent when |y − y0| is small enough. Byhypothesis

|eϕy(x) − eϕy0(x)| 6 M |y − y0|

γ .

This inequality implies

ϕy(x) 6 ϕy0(x) + log(1 +M |y − y0|

γe−ϕy0(x)

)

ϕy(x) > ϕy0(x) + log(1 −M |y − y0|

γe−ϕy0(x)

).

In particular, for ϕy0(x) = r , we have (1 − ε)ϕy0(x) + εr − ε/2 = r − ε/2 , thus

ϕy(x) > r + log(1 −M |y − y0|γe−r) .

Similarly, for ϕy0(x) = r−1 , we have (1− ε)ϕy0(x)+ εr− ε/2 = r−1+ ε/2 , thus

ϕy(x) 6 r − 1 + log(1 +M |y − y0|γe1−r).

The definition of ψ is thus coherent as soon as M |y − y0|γe1−r 6 ε/2 , i.e.

γ log |y − y0| + log2eM

ε6 r .

In this case ψ coincides with ϕy on a neighborhood of ψ = r , and with

(1 − ε)ϕy0(x) + εr − ε/2

on a neighborhood of the polar set ψ−1(−∞) . By Stokes’ formula applied toν(T, ψ, r) , we infer

ν(T, ϕy , r) = ν(T, ψ, r) > ν(T, ψ) = (1 − ε)pν(T, ϕy0).

From (5.10) we get V (y, r) 6 rν(T, ϕy, r) , hence

Ua(y) 6 ρ(y) + V (y, r)− ar 6 ρ(y) + r(ν(T, ϕy, r) − a

),

Ua(y) 6 ρ(y) + r((1 − ε)pν(T, ϕy0) − a

).(5.13)

Suppose γ log |y − y0| + log(2eM/ε) 6 r0 , i.e. |y − y0| 6 (ε/2eM)1/γer0/γ ;one can then choose r = γ log |y − y0| + log(2eM/ε) , and by (5.13) this yieldsthe inequality asserted in theorem 5.12 .

• Fourth step : application of Hormander’s L2 estimates.

The end of the proof rests upon the following crucial result, known as theHormander-Bombieri-Skoda theorem (cf. [Ho] , [Bo] and [Sk]).

Theorem 5.14. — Let u be a psh function on a pseudoconvex open setΩ ⊂ Cn . For every point z0 ∈ Ω such that e−u is integrable in a neighborhoodof z0 , there exists a holomorphic function F on Ω such that F (z0) = 1 and

∫

Ω

|F (z)|2e−u(z)

(1 + |z|2)n+εdλ(z) < +∞ .

Corollary 5.15. — Let u be a psh function on a complex manifold Y. Theset of points in a neighborhood of which e−u is not integrable is an analytic subsetof Y.

Proof. — The result is local, so we may assume that Y is a ball in Cn. Then the

set of non integrability points of e−u is the intersection of all hypersurfaces F−1(0)defined by the holomorphic functions F such that

∫Y|F |2e−udλ < +∞. Indeed F

must vanish at any non integrability point, and on the other hand theorem 5.14shows that one can choose F (z0) = 1 at any integrability point z0.

The main idea in what follows is due to Kiselman [Ki2]. For all real numbersa, b > 0, we let Za,b be the set of points in a neighborhood of which exp(−Ua/b)is not integrable. Then Za,b is analytic, and as the family (Ua) is increasing in a,we have Za′,b′ ⊃ Za′′,b′′ if a′ 6 a′′, b′ 6 b′′.

Let y0 ∈ Y be a given point. If y0 /∈ Ec, then ν(T, ϕy0) < c by definition of Ec.Choose a such that ν(T, ϕy0) < a < c. Lemma 5.9 (a) implies that Ua is boundedbelow in a neighborhood of y0, thus exp(−Ua/b) is integrable and y0 /∈ Za,b for allb > 0.

On the other hand, if y0 ∈ Ec and if a < c, then lemma 5.12 implies for allε > 0 that

Ua(y) 6 (1 − ε)(c− a)γ log |y − y0| + C(ε)

on a neighborhood of y0. Hence exp(−Ua/b) is non integrable at y0 as soon asb < (c− a)γ/2n′, where n′ = dimY. We obtain therefore

Ec =⋂

a<cb<(c−a)γ/2n′

Za,b.

This proves that Ec is an analytic subset of Y .

Exercise 5.16. — Combine El Mir’s extension theorem and Siu’s theorem inorder to get the following result : let P be a complete pluripolar set in a complexmanifold X and A an analytic subset of X \ P. If A has finite mass near everypoint of P, then A is analytic in X.

This result has been first obtained by E. Bishop when P is an analytic subsetof X. The general case is due to Siu.

6. Monge-Ampere capacities and quasi-continuity.

Let Ω be a bounded open subset of Cn. We denote by P (Ω) the set of allpsh functions that are 6≡ −∞ on each connected component of Ω. The followingfundamental definition has been introduced in [B–T2].

Definition 6.1. — For every Borel subset E ⊂ Ω, we set

c(E,Ω) = sup ∫

E

(ddcu)n ; u ∈ P (Ω), 0 6 u 6 1.

The Chern-Levine-Nirenberg inequalities show that c(E,Ω) < +∞ as soon asE ⊂⊂ Ω. If Ω ⊂ B(z0, R), we can choose u(z) = R−2|z − z0|

2 and we obtain

therefore

c(E,Ω) >2nn!

πnR2nλ(E)

where λ is the Lebesgue measure. From the standard properties of measures(countable additivity, monotone convergence theorem), we immediately deduce :

Properties 6.2. — Let Ω ⊂⊂ Cn and E,E1, E2, . . . Borel subsets of Ω.Denote c(E) = c(E,Ω) for simplicity.

(a) If E1 ⊂ E2 then c(E1) 6 c(E2).

(b) c(⋃j>1 Ej) 6

∑j>1 c(Ej).

(c) If E1 ⊂ E2 ⊂ . . . then c(⋃Ej) 6 limj→+∞ c(Ej).

A set function c : E 7→ c(E) defined on all Borel subsets E ⊂ Ω with values in[0,+∞] is called a capacity, resp. a subadditive capacity, if c satisfies the axioms(a,c), resp. (a,b,c) and c(∅) = 0. The capacity is said to be inner regular if all Borelsubsets satisfy

(d) c(E) = supK compact⊂E

c(K).

Similarly, c is said to be outer regular if all Borel subsets E satisfy

(e) c(E) = infG open ⊃E

c(G)

Example 6.3. — If (µα) is a family of positive Radon measures on Ω, thenc(E) = supµα(E) is a subadditive capacity. In general, c does not satisfy theadditivity property

E1, E2 disjoint ⇒ c(E1 ∪ E2) = c(E1) + c(E2) ;

for a specific example, consider the measures µ1 = δ0, µ2 = dλ on R and the setsE1 = 0, E2 =]0, 1] ; then

c(0) = 1 , c(]0, 1]) = 1 , c([0, 1]) = 1.

Moreover, the capacity c = supµα is inner regular because all Radon measures ona separable locally compact space are inner regular. However, c need not be outerregular : for instance, take dµα(x) = α−1ρ(x/α)dx on R, α > 0, where ρ > 0 isa function with support in [−1, 1] and

∫Rρ(x)dx = 1 ; then c(0) = 0 but every

neighborhood of 0 has capacity 1.

The capacity c(•,Ω) defined in 6.1 is called the relative Monge-Ampere capacityon Ω. It is associated to the family of measures µu = (ddcu)n, u ∈ P (Ω), 0 6 u 6 1.In particular c(•,Ω) is inner regular. It is also outer regular, but this fact is nontrivial and will be proved only in § 9.

When c is a capacity and E ⊂ Ω an arbitrary subset, we define the innercapacity c⋆(E) and outer capacity c⋆(E) by

(6.4′) c⋆(E) = supK compact⊂E

c(K),

(6.4′′) c⋆(E) = infG open⊃E

c(G),

A set E ⊂ Ω is said to be c-capacitable if c⋆(E) = c⋆(E). By definition, c is thusregular if and only if all Borel subsets are c-capacitable.

Now we compare capacities associated to different open sets Ω.

Theorem 6.5. — Let Ω1 ⊂ Ω2 ⊂⊂ Cn. Then

(a) c(E,Ω1) > c(E,Ω2) for all Borel subsets E ⊂ Ω1.

(b) Let ω ⊂⊂ Ω1. There exists a constant A > 0 such that c(E,Ω1) 6 Ac(E,Ω2)for all Borel subsets E ⊂ ω.

Proof. — Since every psh function u ∈ P (Ω2) with 0 6 u 6 1 induces a pshfunction in P (Ω1) with the same property, (a) is clear.

(b) Use a finite covering of ω by open balls contained in Ω1 and cut E intopieces. The proof is then reduced to the case when ω ⊂⊂ Ω1 are concentric balls,say Ω1 = B(0, r) and ω = B(0, r − ε). For every u ∈ P (Ω1) such that 0 6 u 6 1,set

u(z) =

maxu(z), λ(|z|2 − r2) + 2 on Ω1,λ(|z|2 − r2) + 2 on Ω2 \ Ω1.

Choose λ so large that λ((r − ε)2 − r2) 6 −2. Then u ∈ P (Ω2) and u = u on ω.Moreover 0 6 u 6 M for some constant M > 0, thus for E ⊂ ω we get

∫

E

(ddcu)n =

∫

E

(ddcu)n 6 Mnc(E,Ω2).

Therefore c(E,Ω1) 6 Mnc(E,Ω2).

As a consequence of theorem 6.5, it is in general harmless to shrink the domainΩ when one wants to estimate capacities.

Theorem 6.6. — Let K be a compact subset of Ω and ω ⊂⊂ Ω a neighbor-hood of K. There is a constant A > 0 such that for every v ∈ P (Ω)

c(K ∩ v < −m,Ω) 6 A‖v‖L1(ω) ·1

m.

Proof. — For every u ∈ P (Ω), 0 6 u 6 1, theorem 1.8 implies∫

K∩v<−m

(ddcu)n 61

m

∫

K

|v|(ddcu)n 61

mCK,ω‖v‖L1(ω).

Definition 6.7. — A set P ⊂ Ω is said to be pluripolar in Ω if there existsv ∈ P (Ω) such that P ⊂ v = −∞.

Corollary 6.8. — If P is pluripolar in Ω, then

c⋆(P,Ω) = 0.

Proof. — Write P ⊂ v = −∞ and Ω =⋃j>1 Ωj with Ωj ⊂⊂ Ω. Theorem 6.4

shows that there is an open set Gj = Ωj ∩ v < −mj such that c(Gj ,Ω) < ε2−j .Then v = −∞ ⊂ G =

⋃Gj and c(G,Ω) < ε.

Theorem 6.9. — Let vk, v ∈ P (Ω) be locally bounded psh functions suchthat (vk) decreases to v. Then for every compact subset K ⊂ Ω and every δ > 0

limk→+∞

c(K ∩ vk > v + δ,Ω) = 0.

Proof. — It is sufficient to show that

supu∈P (Ω),06u61

∫

K

(vk − v)(ddcu)n

tends to 0, because this supremum is larger than δ.c(K ∩ vk > v + δ,Ω). Bycutting K into pieces and modifying v, vk, u with the max construction, we mayassume that K ⊂ Ω = B(0, r) are concentric balls and that all functions v, vk, uare equal to λ(|z|2 − r2) + 2) on the corona Ω \ω, ω = B(0, r− ε). An integrationby parts yields

∫

Ω

(vk − v)(ddcu)n = −

∫

Ω

d(vk − v) ∧ dcu ∧ (ddcu)n−1.

The Cauchy-Schwarz inequality implies that this integral is bounded by

A(∫

Ω

d(vk − v) ∧ dc(vk − v) ∧ (ddcu)n−1)1/2

where

A2 =

∫

Ω

du ∧ dcu ∧ (ddcu)n−1 6

∫

Ω

ddc(u2) ∧ (ddcu)n−1

and this last integral depends only on the constants λ, r. Another integration byparts yields

∫

Ω

d(vk − v) ∧ dc(vk − v) ∧ (ddcu)n−1 = −

∫

Ω

(vk − v)ddc(vk − v) ∧ (ddcu)n−1

6

∫

Ω

(vk − v)ddcv ∧ (ddcu)n−1.

We have thus replaced one factor ddcu by ddcv in the integral. Repeating theargument (n− 1) times we get

∫

Ω

(vk − v)(ddcu)n 6 C( ∫

Ω

(vk − v)(ddcv)n)1/2n

and the last integral converges to 0 by the bounded convergence theorem.

Theorem 6.10 (Quasi-consinuity of psh functions). — Let Ω ⊂⊂ Cn andv ∈ P (Ω). Then for each ε > 0, there is an open subset G of Ω such that c(G,Ω) < εand v is continuous on Ω \G.

Proof. — Let ω ⊂⊂ Ω be arbitrary. We first show that there exists G ⊂ ωsuch that c(G,Ω) < ε and v continuous on ω \ G. For m > 0 large enough, theset G0 = ω ∩ v < −m has capacity < ε/2 by theorem 6.6. On ω \ G0 we havev > −m, thus v = maxv,−m coincides with v there and v is locally boundedon Ω. Let (vk) be a sequence of smooth psh functions which decrease to v in aneighborhood of ω. For each ℓ > 1, theorem 6.9 shows that there is an index k(ℓ)and an open set

Gk(ℓ) = ω ∩ vk(ℓ) > v + 1/ℓ

such that c(Gk(ℓ),Ω) < ε2−ℓ−1. Then G = G0 ∪⋃Gk(ℓ) has capacity c(G,Ω) < ε

by 6.2 (b) and (vk(ℓ)) converges uniformly to v = v on ω \G. Hence v is continuouson ω \ G. Now, take an increasing sequence ω1 ⊂ ω2 ⊂ . . . with

⋃ωj = Ω and

Gj ⊂ ωj such that c(Gj ,Ω) < ε2−j and v continuous on ωj \Gj . The set G =⋃Gj

satisfies all requirements.

As an example of application, we prove an interesting inequality for the Monge-Ampere operator.

Proposition 6.11. — Let u, v be locally bounded psh functions on Ω. Thenwe have an inequality of measures

(ddcmaxu, v)n > 1u>v(ddcu)n + 1u<v(dd

cv)n.

Proof. — It is enough to check that∫

K

(ddcmaxu, v)n >

∫

K

(ddcu)n

for every compact set K ⊂ u > v ; the other term is then obtained by reversingthe roles of u and v. By shrinking Ω, adding and multiplying with constants,we may assume that 0 6 u, v 6 1 and that u, v have regularizations uε, vε with0 6 uε, vε 6 1 on Ω. Let G ⊂ Ω be an open set of small capacity such that u, vare continuous on Ω \ G. By Dini’s lemma, uε, vε converge uniformly to u, v onΩ \G. Hence for any δ > 0, we can find an arbitrarily small neighborhood L of Ksuch that uε > vε − δ on L \G for ε small enough. As (ddcuε)

n converges weaklyto (ddcu)n on Ω, we get

∫

K

(ddcu)n 6 lim infε→0

∫

L

(ddcuε)n

6 lim infε→0

( ∫

G

(ddcuε)n +

∫

L\G

(ddcuε)n)

6 c(G,Ω) + lim infε→0

∫

L\G

(ddcmaxuε + δ, vε)n.

Observe that maxuε + δ, vε coincides with uε + δ on a neighborhood of L \ G.By weak convergence again, we get

∫

K

(ddcu)n 6 c(G,Ω) +

∫

L\G

(ddcmaxu+ δ, v)n.

By taking L very close to K and c(G,Ω) arbitrarily small, this implies∫

K

(ddcu)n 6

∫

K

(ddcmaxu+ δ, v)n

and the desired conclusion follows by letting δ tend to 0.

7. Extremal functions and negligible sets.

Let (uα) be a family of upper semi-continuous functions on Ω which is locallybounded from above. Then the upper envelope

u = supαuα(z)

need not be upper semi-continuous, so we consider its “upper semi-continuousregularization”

u⋆(z) = limε→0

supB(z,ε)

u > u(z).

It is easy to check that u⋆ is upper semi-continuous and that u⋆ is the smallestupper semi-continuous function > u.

Let B(zj, εj) be a denumerable basis of the topology of Ω. For each j, let (zjk)be a sequence in B(zj, εj) such that

supku(zjk) = sup

B(zj ,εj)

u,

and for each (j, k), let α(j, k, ℓ) be a sequence of indices α such that u(zjk) =supℓ uα(j,k,ℓ)(zjk). Set

v = supj,k,ℓ

uα(j,k,ℓ).

Then v 6 u and v⋆ 6 u⋆. On the other hand

supB(zj ,εj)

v > supkv(zjk) > sup

k,ℓuα(j,k,ℓ)(zjk) = sup

ku(zjk) = sup

B(zj ,εj)

u.

As every ball B(z, ε) is a union of balls B(zj, εj), we easily conclude that v⋆ > u⋆,hence v⋆ = u⋆. Therefore :

Choquet’s lemma 7.1. — Every family (uα) has a denumerable subfamily(uα(j)) whose upper envelope v satisfies v 6 u 6 u⋆ = v⋆.

Proposition 7.2. — If all uα are psh, then u⋆ is psh and equal almosteverywhere to u.

Proof. — By Choquet’s lemma one may assume that (uα) is denumerable.Then u = supuα is a Borel function. For every (z0, a) ∈ Ω × Cn, uα satisfies themean value inequality on circles, hence

u(z0) = sup uα(z0) 6 sup

∫ 2π

0

uα(z0 + aeiθ)dθ

2π6

∫ 2π

0

u(z0 + aeiθ)dθ

2π.

It follows easily that each convolution u⋆ρε also satisfies the mean value inequality,thus u ⋆ ρε is smooth and psh. Therefore (u ⋆ ρε) ⋆ ρη is increasing in η. Lettingε tends to 0, we see that u ⋆ ρη in increasing in η. Since u ⋆ ρε is smooth andu ⋆ ρε > u by the mean value inequality, we also have u ⋆ ρε > u⋆. By the uppersemi-continuity we get limε→0 u ⋆ ρε = u⋆, in particular u⋆ is psh and coincidesalmost everywhere with the L1

loc limit u.

A set of the form

(7.3) N = z ∈ Ω ; u(z) < u⋆(z)

is called negligible. Every pluripolar set P = v = −∞ is negligible : letw ∈ P (Ω) ∩ C∞(Ω) such that w > v and uα = (1 − α)v + αw, α ∈]0, 1[. Then uαis increasing in α and u = supα uα satisfies

u = −∞ on v = −∞,u = w on v > −∞.

Hence u⋆ = w and u < u⋆ = v = −∞.

To study further properties of the capacity, we consider the extremal functionassociated to a subset E of Ω :

(7.4) uE(z) = supv(z); v ∈ P (Ω), v 6 −1 on E, v 6 0 on Ω.

Proposition 7.2 implies u⋆E ∈ P (Ω) and −1 6 u⋆E 6 0. In the sequel, we needthe fundamental result of Bedford-Taylor [B–T1] on the solution of the Dirichletproblem for complex Monge-Ampere equations.

Theorem 7.5. — Let Ω ⊂⊂ Cn be a smooth strongly pseudoconvex domainand let f ∈ C0(∂Ω) be a continuous function on the boundary. Then

u(z) = supv(z) ; v ∈ P (Ω) ∩ C0(Ω), v 6 f on ∂Ω

is continuous on Ω and psh on Ω, and solves the Dirichlet problem

(ddcu)n = 0 on Ω , u = f on ∂Ω.

Sketch of proof. — Let g ∈ C2(Ω) be an approximate extension of f suchthat |g − f | < ε on ∂Ω and let ψ < 0 be a smooth strongly psh exhaustion of Ω.Then g − ε + Aψ is psh for A > 0 large enough and g − ε + Aψ = g − ε 6 f on∂Ω, hence g − ε + Aψ 6 u on Ω. Similarly, for all v ∈ P (Ω) ∩ C0(Ω) with v 6 fon ∂Ω, the function v − g − ε+ Aψ equals v − g − ε 6 0 on ∂Ω and is psh for Alarge, thus v − g − ε+Aψ 6 0 on Ω by the maximum principle. Therefore we getu 6 g+ε−Aψ ; as ε tends to 0, we see that u = f on ∂Ω and that u is continuousat every point of ∂Ω. Since g+ ε+Aψ = g+ ε > f on ∂Ω, there exists δ > 0 suchthat u⋆ < g + ε + Aψ on Ω \ Ωδ, where Ωδ = ψ < −δ. For η > 0 small enough,the regularizations of u⋆ satisfy u⋆ ⋆ ρη < g + ε + Aψ on a neighborhood of ∂Ωδ.Then we let

vε =

maxu⋆ ⋆ ρη − 2ε, g − ε+Aψ on Ωδg − ε+ Aψ on Ω \ Ωδ.

It is clear that vε is psh and continuous on Ω and that vε = g−ε 6 f on ∂Ω, hencevε 6 u on Ω. We obtain therefore u⋆ ⋆ ρη 6 u+ 2ε on Ωδ. As u 6 u⋆ 6 u⋆ ⋆ ρη, wesee that u⋆ ⋆ ρη converges uniformly to u on every compact subset of Ω. Hence uis psh and continuous on Ω.

We shall complete the proof under the following additional assumptions :f ∈ C2(∂Ω) and u ∈ C2(Ω) ; the general case is difficult and rather technical(cf. [B–T1]). The plurisubharmonicity of u implies det(∂2u/∂zj∂zk) > 0. If wehad a strict inequality at one point z0 ∈ Ω, say z0 = 0 for simplicity, the Taylorexpansion of u at z0 would give

u(z) = ReP (z) +∑

cjkzjzk + o(|z|2)

where P is a holomorphic polynomial of degree 2 and (cjk) a positive definitehermitian matrix. Hence we would have u > ReP + ε on a small sphere S(0, r)with B(0, r) ⊂ Ω. The function

v =

maxu,ReP + ε on B(0, r)u on Ω \B(0, r)

is then continuous on Ω and psh, and satisfies

v = u 6 f on ∂Ω.

By the definition of u, we thus have u > v on Ω. This is a contradiction, because

v(0) > ReP (0) = u(0).

Corollary 7.6. — Fix a ball B(z0, r) ⊂ Ω and let f ∈ P (Ω) be locallybounded. There exists a function f ∈ P (Ω) such that f > f on Ω, f = f onΩ \B(0, r) and (ddcf)n = 0 on B(z0, r). Moreover, for f1 6 f2 we have f1 6 f2.

Proof. — Assume first that f ∈ C0(Ω). By theorem 7.5 applied on B(z0, r),there exists u psh and continuous on B(z0, r) with u = f on S(z0, r) and(ddcu)n = 0 on B(z0, r). Set

f =

u on B(z0, r)f on Ω \B(z0, r).

By definition of u, we have f = u > f on B(z0, r). Moreover, f is the decreasinglimit of the psh functions

gk =

maxu, f + 1

k on B(z0, r)f + 1

knear Ω \B(z0, r)

hence f is psh. Also clearly, for f1 6 f2 we have u1 6 u2, hence f1 6 f2. Foran arbitrary locally bounded function f ∈ P (Ω), write f as a decreasing limit ofsmooth psh functions fk = f ⋆ ρ1/k and set f = limk→+∞ ↓ fk. Then f has allrequired properties.

Now we prove the following three fundamental results by a simultaneousinduction on n.

Proposition 7.7. — Let u, uj ∈ P (Ω) be locally bounded functions suchthat uj increases to u almost everywhere. Then the measure (ddcuj)

n convergesweakly to (ddcu)n on Ω.

Proposition 7.8. — Let Ω be a strongly pseudoconvex smooth open subsetof Cn. If K ⊂ Ω is compact, then

(a) (ddcu⋆K)n = 0 on Ω \K.

(b) c(K,Ω) =∫K

(ddcu⋆K)n =∫Ω(ddcu⋆K)n.

Proposition 7.9. — If a Borel set N ⊂ Ω is negligible, then

c(N,Ω) = 0.

The inductive proof is made in three steps.

Step 1 : 7.7 in Cn ⇒ 7.8 in Cn.Step 2 : 7.8 in C

n ⇒ 7.9 in Cn.

Step 3 : 7.7 and 7.9 in Cn ⇒ 7.7 in Cn+1.

In the case n = 1, proposition 7.7 is a well-known fact of distribution theory : ujconverges to u in L1

loc(Ω), thus ddcuj converges weakly to ddcu. By the inductiveargument, propositions 7.7, 7.8, 7.9 hold in all dimensions.

Proof of step 1. — By Choquet’s lemma, there is a sequence of functionsvj ∈ P (Ω) such that vj 6 0 on Ω, vj 6 −1 on K and v⋆ = u⋆K . If we replacevj by max−1, v1, . . . , vj, we see that we may assume vj > −1 for all j and vjincreasing. Then fix an arbitrary ball B(z0, r) ⊂ Ω\K and consider the increasingsequence vj given by corollary 7.6. We still have vj 6 0 on Ω and vj 6 −1 onK, thus vj 6 vj 6 uK and v = lim vj satisfies v⋆ = v⋆ = u⋆K , in particularlim vj = lim vj = u⋆K almost everywhere. Since (ddcvj)

n = 0 on B(z0, r), weconclude by 7.7 that (ddcu⋆K)n = 0 on B(z0, r) and 7.8 (a) is proved.

To prove 7.8 (b), observe first that −1 6 u⋆K 6 0 on Ω, hence c(K,Ω) >∫K

(ddcu⋆K)n by definition of the capacity. If ψ < 0 is a smooth strictly pshexhaustion function of Ω, we have Aψ 6 −1 on K for A large enough. We canclearly assume vj > Aψ on Ω ; otherwise replace vj by maxvj , Aψ. Now, letw ∈ P (Ω) be such that 0 6 w 6 1 and set

w′ = (1 − ε)w − 1 + ε/2 , wj = maxw′, vj.

Since −1 + ε/2 6 w′ 6 −ε/2 on Ω, we have wj = vj as soon as Aψ > −ε/2,whereas wj = w′ > −1 + ε/2 > vj on a neighborhood of K. Hence for δ > 0 smallenough Stokes’ theorem implies∫

Ωδ

(ddcvj)n =

∫

Ωδ

(ddcwj)n >

∫

K

(ddcwj)n = (1 − ε)n

∫

K

(ddcw)n.

By 7.7 (ddcvj)n converges weakly to (ddcu⋆K)n and we get

lim supj→+∞

∫

Ωδ

(ddcvj)n 6

∫

Ωδ

(ddcu⋆K)n =

∫

K

(ddcu⋆K)n.

Therefore∫K

(ddcw)n 6∫K

(ddcu⋆K)n and c(K,Ω) 6∫K

(ddcu⋆K)n.

Proof of step 2. — Let N = v < v⋆ with v = sup vα. By Choquet’s lemma,we may assume that vα is an increasing sequence of psh functions. The theorem ofquasi-continuity shows that there exists an open set G ⊂ Ω such that all functionsvα and v⋆ are continuous on Ω \G and c(G,Ω) < ε. Write

N ⊂ G ∪ (N ∩ (Ω \G)) = G ∪⋃

δ,λ,µ∈Q

Kδλµ

where δ > 0, λ < µ and

Kδλµ =z ∈ Ωδ \G ; v(z) 6 λ < µ 6 v⋆(z)

.

As v⋆ is continuous and v lower semi-continuous on Ω \ G, we see that Kδλµ iscompact. We only have to prove that c(Kδλµ,Ω) = 0. Set K = Kδλµ for simplicityand take an open set ω ⊂⊂ Ω. By subtracting a large constant, we may assumev⋆ 6 0 on ω.

Multiplying by another constant, we may set λ = −1. Then all vα satisfy vα 6 0on ω and vα 6 v 6 −1 on K. We infer that the extremal function uK on ω satisfiesuK > v, u⋆K > v⋆, in particular u⋆K > µ > −1 on K. By proposition 6.11 we obtain

c(K,ω) =

∫

K

(ddcu⋆K)n 6

∫

K

(ddcmaxu⋆K , µ)n 6 |µ|nc(K,ω)

because −1 6 |µ|−1 maxu⋆K , µ 6 0. As |µ| < 1, we conclude that c(K,ω) = 0,hence c(K,Ω) = 0.

Proof of step 3. — We have to show that if Ω ⊂ Cn+1,

limj→+∞

∫

Ω

χ(ddcuj)n+1 =

∫

Ω

χ(ddcu)n+1

for all test functions χ ∈ C∞0 (Ω). That is,

(7.10) limj→+∞

∫

Ω

uj(ddcuj)

n+1 ∧ γ =

∫

Ω

u(ddcu)n ∧ γ

with γ = ddcχ. As all (1, 1)-forms γ can be written as linear combinations of formsof the type iα∧ α, α ∈ Λ1,0(Cn)⋆, it is sufficient, after a change of coordinates, toconsider forms of the type γ = i

2χ(z)dzn+1 ∧ dzn+1, χ ∈ C∞0 (Ω). In this case, for

any locally bounded psh function v on Ω, the Fubini theorem yields∫

Ω

v(ddcv)n ∧ γ =

∫

C

dλ(zn+1)

∫

Ω(zn+1)

χ(•, zn+1)(ddcv(•, zn+1))

n

where Ω(zn+1) = z ∈ Cn; (z, zn+1) ∈ Ω and f(•, zn+1) denotes the function

z 7→ f(z, zn+1) on Ω(zn+1). Indeed, the result is clearly true if v is smooth.The general case follows by taking smooth psh functions vj decreasing to v.The convergence of both terms in the equality is guaranteed by theorem 1.5 (a),combined with 1.3 and the bounded convergence theorem for the right hand side.

In order to prove (7.10), we thus have to show

(7.11) limj→+∞

∫

ω

χuj(ddcuj)

n =

∫

ω

χu(ddcu)n

for ω ⊂ Cn, χ ∈ C∞0 (ω) and uj ∈ P (ω) ∩ L∞

loc(ω) increasing to u ∈ P (ω) almosteverywhere. To prove (7.11), we can clearly assume 0 6 χ 6 1 and 0 6 uj 6 u 6 1on Ω. By our inductive hypothesis 7.7, (ddcuj)

n converges weakly to (ddcu)n. Asuj 6 u 6 uε = u ⋆ ρε, we get

lim supj→+∞

∫

ω

χuj(ddcuj)

n 6 limε→0

limj→+∞

∫

ω

χuε(ddcuj)

n

= limε→0

∫

ω

χuε(ddcu)n =

∫

ω

χu(ddcu)n.

To prove the other inequality, let ε > 0 and choose an open set G ⊂ ω such thatc(G, ω) < ε and u, uj are all continuous on ω \ G. Let v = sup uj . Then v⋆ = ubecause v⋆ and u are psh and coincide almost everywhere. Let uj be a continuousextension of uj |ω\G to ω such that 0 6 uj 6 1. For j > k we have uj > uk, hence

∫

ω

χuj(ddcuj)

n >

∫

ω\G

χuk(ddcuj)

n

>

∫

ω

χuk(ddcuj)

n −

∫

G

(ddcuj)n.

The last integral on the right is 6 c(G, ω) < ε. Taking the limit as j tends to +∞,we obtain

lim infj→+∞

∫

ω

χuj(ddcuj)

n >

∫

ω

χuk(ddcu)n − ε

>

∫

ω

χuk(ddcu)n − 2ε.

The second term ε comes from∫G

(ddcu)n 6 c(G, ω) < ε. Now let k → +∞ andε→ 0 to get

lim infj→+∞

∫

ω

χuj(ddcuj)

n >

∫

ω

χv(ddcu)n.

Moreover, the Borel set N = v < u = v⋆ is negligible and the inductivehypothesis 7.9 implies c(N,ω) = 0. Therefore

∫

ω

χ(u− v)(ddcu)n 6

∫

N

(ddcu)n = 0

and the proof is complete.

Theorem 7.12. — For each j = 1, . . . , q, let ukj be an increasing sequence

of locally bounded psh functions such that ukj converges almost everywhere touj ∈ P (Ω). Then

(a) ddcuk1 ∧ . . . ∧ ddcukq → ddcu1 ∧ . . . ∧ ddcuq weakly.

(b) uk1ddcuk2 ∧ . . . ∧ ddcukq → u1dd

cu2 ∧ . . . ∧ ddcuq weakly.

Proof. — (a) Without loss of generality, we may assume q = n, otherwise wecomplete with additional stationary sequences ukq+1 = uq+1, . . . , u

kn = un where

uq+1, . . . , un are chosen arbitrarily in P (Ω) ∩ C∞(Ω). Now apply proposition 7.7to uk = λ1u

k1 + · · · + λnu

kn, λj > 0, and consider the coefficient of λ1 . . . λn in

(ddcuk)n.

(b) Same proof as for (7.11).

Exercise 7.13. — Use formula 7.8 (b) in order to compute the capacity ofK = B(0, r) in a ball Ω = B(0, R) ; show that the extremal function uK is

uK(z) =(

logR

r

)−1

max

log|z|

R, log

r

R

.

8. Characterization of pluripolar sets.

First we quote a few elementary properties of the extremal functions u⋆E .

Properties 8.1. —

(a) if E1 ⊂ E2 ⊂ Ω, then u⋆E1> u⋆E2

.

(b) if E ⊂ Ω1 ⊂ Ω2, then u⋆E,Ω1> u⋆E,Ω2

.

(c) if E ⊂ Ω, then u⋆E = uE = −1 on E0 and (ddcu⋆E)n = 0 on Ω \ E ; hence(ddcu⋆E)n is supported by ∂E.

(d) u⋆E = 0 if and only if there exists v ∈ P (Ω), v 6 0 such that E ⊂ v = −∞.

(e) if E ⊂⊂ Ω and if Ω is strongly pseudoconvex with exhaustion ψ < 0, thenu⋆E > Aψ for some A > 0.

Proof. — (a), (b) are obvious from definition (7.4); (e) is true as soon asAψ 6 −1 on E ; the equality (ddcu⋆E)n = 0 on Ω \ E in (c) is proved exactly inthe same way as 7.8 (a) in step 1.

(d) If E ⊂ v = −∞, v ∈ P (Ω), v 6 0, then for every ε > 0 we have εv 6 uE ,hence uE = 0 on Ω \ v = −∞ and u⋆E = 0.

Conversely, Choquet’s lemma shows that there is an increasing sequencevj ∈ P (Ω), −1 6 vj 6 uE , converging almost everywhere to u⋆E . If u⋆E = 0,we can extract a subsequence in such a way that

∫Ω|vj |dλ < 2−j . As vj 6 0 and

vj 6 −1 on E, the function v =∑vj is psh 6 0 and v = −∞ on E.

If G ⊂⊂ Ω is an open subset, K1 ⊂ K2 ⊂ . . . compact subsets of G withKj ⊂ K0

j+1 and⋃Kj = G, then u⋆Kj

= −1 on K0j ⊃ Kj−1, lim ↓ u⋆Kj

= −1 on G.

Therefore u⋆G 6 limu⋆Kj6 uG 6 u⋆G and theorems 1.5, 7.8, 8.1 (c) show that

(8.2) c(G,Ω) =

∫

G

(ddcu⋆G)n =

∫

Ω

(ddcu⋆G)n.

Proposition 8.3. — Let Ω ⊂⊂ Cn and K1 ⊃ K2 ⊃ . . . , K =

⋂Kj compact

subsets of Ω. Then

(a)(

limj→+∞

↑ u⋆Kj

)⋆= u⋆K .

(b) lim c(Kj ,Ω) = c(K,Ω).

(c) c⋆(K,Ω) = c(K,Ω).

Proof. — We have lim ↑ u⋆Kj6 u⋆K by 8.1 (a). On the other hand, let v ∈ P (Ω)

be such that v 6 0 on Ω and v 6 −1 onK. For every ε > 0 the open set v < −1+εis a neighborhood of K, thus K ⊂ v < −1 + ε for j large. We obtain thereforev−ε 6 u⋆Kj

and uK = supv 6 limu⋆Kj, whence equality (a). Property (b) follows

now from theorems 7.7 and 7.8 (b), and (c) is a consequence of (b) when Kj areneighborhoods of K.

Lemma 8.4. — Let Ω ⊂⊂ Cn and u, v ∈ P (Ω) locally bounded psh functionssuch that u 6 v 6 0 and lim

z→∂Ωu(z) = 0. Then

∫

Ω

(ddcv)n 6

∫

Ω

(ddcu)n.

Moreover∫Ω(ddcu)n = 0 if and only if u = 0.

Proof. — As maxu+ ε, v = u+ ε near ∂Ω, we get∫

Ω

(ddcu)n =

∫

Ω

(ddcmaxu+ ε, v)n.

Let ε tend to 0, and observe that the integrand on the right hand side convergesweakly to (ddcv)n by theorem 1.5. The asserted inequality follows.

Now, assume that u(z0) < 0 at some point. Then

v(z) = maxu(z), ε2|z|2 − ε

coincides with u near ∂Ω and with ε2|z|2 − ε on a neighborhood ω of z0. We gettherefore ∫

Ω

(ddcu)n =

∫

Ω

(ddcv)n >

∫

ω

(ddcv)n > 0.

Proposition 8.5. — Let Ω ⊂⊂ Cn be strongly pseudoconvex. If E ⊂⊂ Ω is

an arbitrary subset, then

c⋆(E,Ω) =

∫

Ω

(ddcu⋆E)n.

Proof. — Let ψ < 0 be a strictly psh exhaustion function on Ω. For everyopen set G ⊃ E, we have u⋆E > u⋆G > Aψ by 8.1, and lemma 8.4 implies

∫

Ω

(ddcu⋆E)n 6

∫

Ω

(ddcu⋆G)n = c(G,Ω),

thus∫Ω(ddcu⋆E)n 6 c⋆(E,Ω).

Conversely, Choquet’s lemma shows that there exists an increasing sequencevj ∈ P (Ω) with −1 6 vj 6 0, vj > Aψ on Ω and lim vj = uE almost everywhere.If

Gj = z ∈ Ω; (1 + 1/j)vj(z) < −1

then Gj ⊃ E, Gj is decreasing and (1+1/j)vj 6 uGj. Thus lim ↑ u⋆Gj

= u⋆E almosteverywhere and theorem 7.7 gives

limj→+∞

∫

Ω

(ddcu⋆Gj)n =

∫

Ω

(ddcu⋆E)n.

Corollary 8.6. — Let Ω ⊂⊂ Cn be strongly pseudoconvex. If E ⊂⊂ Ω, thenc⋆(E,Ω) = 0 if and only if u⋆E = 0.

Now, we prove an important result due to Josefson [Jo]. A set P in Cn is saidto be locally pluripolar if for each z ∈ P there is an open neighborhood Ω of z andv ∈ P (Ω) such that P ∩ Ω ⊂ v = −∞.

Theorem 8.7 (Josefson). — If P ⊂ Cn is locally pluripolar, there existsv ∈ P (Cn) with P ⊂ v = −∞, i.e. P is globally pluripolar in Cn.

Proof. — By the definition of locally pluripolar, one can find sets Pj ,Ωj withPj ⊂⊂ Ωj ⊂⊂ Cn,

⋃j>1 Pj = P, each Ωj strongly pseudoconvex. By 8.1 (d) and

8.5, we have c⋆(Pj,Ωj) = 0.

Let Bk be the ball of center 0 and radius k in Cn and j(k) a sequence of integerssuch that each integer is repeated infinitely many times and Ωj(k) ⊂ Bk. By thecomparison theorem 6.5 we have c⋆(Pj(k), Bk+2) = 0, hence the extremal functionu⋆Pj(k)

in Bk+2 is zero and we can find vk ∈ P (Bk+2) with −1 6 vk 6 0, vk = −1

on Pj(k) and∫Bk

|vk|dλ < 2−k. Now set

vk(z) =

vk(z) on Bkmaxvk(z), |z|

2 − (k + 1)2 on Bk+2 \Bk|z|2 − (k + 1)2 on Cn \Bk+2.

As vk 6 0 on Bk and∫Bk

|vk|dλ < 2−k, the series v =∑vk defines a global psh

function on Cn. Moreover vk = −1 on Pj(k) and each Pj is repeated infinitely

many times, therefore

v = −∞ on⋃Pj = P.

Corollary 8.8. — Let Ω ⊂ Cn and P ⊂ Ω. Then P is pluripolar in Ω if

and only if c⋆(P,Ω) = 0.

Proof. — That P is pluripolar implies c⋆(P,Ω) = 0 was proved in corollary6.8. Conversely, if c⋆(P,Ω) = 0 then c⋆(P ∩ ω′, ω) = 0 for all concentric balls

ω′ ⊂⊂ ω ⊂⊂ Ω and corollary 8.6 combined with 8.1 (d) shows that P ∩ ω′ ispluripolar in ω. Josefson’s theorem implies that P is globally pluripolar in Cn.

Corollary 8.9. — Negligible sets are pluripolar.

Proof. — By Choquet’s lemma every negligible set is contained in a Borelnegligible set N = v < v⋆ with v = sup vj . However, in step 2 of § 7, we showedthat N ⊂ G ∪

⋃Kδλµ with G open, c(G,Ω) < ε and c(Kδλµ,Ω) = 0. By 8.3 (c),

we have c⋆(N,Ω) < ε for all ε > 0. Therefore c⋆(N,Ω) = 0 and N is pluripolar.

9. Capacitability and outer regularity.

First we introduce some definitions and prove a general capacitability result dueto G. Choquet. All topological spaces occurring here are assumed to be Hausdorff.

Definition 9.1. — Let X be a topological space.

• A Fσ subset of X is a countable union of closed subsets of X ;

• A Fσδ subset of X is a countable intersection of Fσ subsets of X.

• A space X is said to be a Kσ (resp. Kσδ) space if it is homeomorphic to someFσ (resp. Fσδ) subset of a compact space W.

Properties 9.2. —

(a) Every closed subset F of a Kσδ space X is a Kσδ space.

(b) Every countable disjoint sum∐Xj of Kσδ spaces is a Kσδ space.

(c) Every countable product∏Xj of Kσδ spaces is a Kσδ space.

Proof. — (a) Write X =⋂ℓ>1Gℓ and Gℓ =

⋃m>1Kℓm where Kℓm are closed

subsets of a compact space W . If F is the closure of F in W , we have

F = X ∩ F =⋂

ℓ>1

Gℓ ∩ F , Gℓ ∩ F =⋃

m>1

Kℓm ∩ F .

(b) Let Xj , j > 1, be Kσδ spaces and write for each j

Xj =⋂

ℓ>1

Gjℓ , Gjℓ =⋃

m>1

Kjℓm

where Kjℓm is a closed subspace of a compact space Wj . Then

∐Wj can be

embedded in the compact space W =∏

(Wj∐⋆) via the obvious map that sendsw ∈Wj to (⋆, . . . , ⋆, w, ⋆, . . .) where w is in the j-th position. Now X =

∐Xj can

be written

X =⋂

ℓ>1

Gℓ, Gℓ =⋃

m>1

∐

j>1

Kjℓm.

As Kjℓm is sent onto a closed set by the embedding

∐Wj −→ W , we conclude that

X is a Kσδ space.

(c) With the notations of (b), write X =∏Xj as

X =⋂

ℓ>1

Gℓ, Gℓ = G1ℓ ×G2

ℓ−1 × . . .×Gℓ1 ×Wℓ+1 × . . .×Wj × . . . ,

Gℓ =⋃

m1,...,mℓ>1

K1ℓm1

×K2ℓ−1m2

× . . .×Kℓ1mℓ

×Wℓ+1 × . . .×Wj × . . .

where each term in the union is closed in W =∏Wj .

Definition 9.3. — A space E is said to be K-analytic if E is a continuousimage of a Kσδ space X.

Theorem 9.4. — Let Ω be a topological space and E1 ⊂ E2 ⊂ . . . K-analyticsubsets of Ω. Then

⋃Ej and

⋂Ej are K-analytic.

Proof. — Let fj : Xj → Ej be a continuous map from a Kσδ space onto Ej .Set X =

∐Xj and f =

∐fj : X → Ω. Then X is a Kσδ space, f is continuous

and f(X) =⋃Ej. Now set

X = x = (x1, x2, . . .) ∈∏

Xj ; f1(x1) = f2(x2) = · · ·,

f : X → Ω , f(x) = f1(x1) = f2(x2) = · · · .

Then X is closed in∏Xj , so X is a Kσδ space by 9.2 (a,c) and f(X) =

⋂Ej.

Corollary 9.5. — Let Ω be a separable locally compact space. Then allBorel subsets of Ω are K-analytic.

Proof. — Any open or closed open set in Ω is a countable union of compactsubsets, hence K-analytic. On the other hand, theorem 9.4 shows that

A = E ⊂ Ω ; E and Ω \ E are K − analytic

is a σ-algebra. Since A contains all open sets in E, A must also contain all Borelsubsets.

Before going further, we need a simple lemma.

Lemma 9.6. — Let E be a relatively compact K-analytic subset of a topo-logical space Ω. There exists a compact space T, a continuous map g : T → Ω anda Fσδ subset Y ⊂ T such that g(Y ) = E.

Proof. — There is a compact space W, a Fσδ subset X ⊂W and a continuousmap f : X → E onto E. Let

Y = (x, f(x)) ; x ∈ X ⊂ X ×E

be the graph of f and T = Y the closure of Y in the compact space X×E. As f iscontinuous, Y is closed in X ×E, thus Y = T ∩ (X ×E). Now, X is a Fσδ subset

of X , so X ×E is a Fσδ subset of X ×E and Y is a Fσδ subset of T . Finally E isthe image of Y by the second projection g : T → E.

Definition 9.7. — Let Ω be a topological space. A generalized capacity is aset function c defined on all subsets E ⊂ Ω satisfying the following axioms :

(a) If E1 ⊂ E2 ⊂ Ω, then c(E1) 6 c(E2).

(b) If E1 ⊂ E2 ⊂ . . . ⊂ Ω, then c(⋃Ej) = limj→+∞ c(Ej).

(c) If K1 ⊃ K2 ⊃ . . . are compact subsets of Ω, then c(⋂Kj) = limj→+∞ c(Kj).

Choquet’s capacitability theorem 9.8. — Let Ω be a Kσ space and letc be a generalized capacity on Ω. Then every K-analytic subset E ⊂ Ω satisfies

c(E) = supK compact⊂E

c(K).

Proof. — As Ω is an increasing union of compact sets Lj , we have c(E) =limj→+∞ c(E ∩ Lj) by axiom (b); we may therefore assume that E is relativelycompact in Ω. Then lemma 9.6 shows that there is a Fσδ subset Y in a compactspace T and a continuous map g : T → Ω such that g(Y ) = E. It is immediate tocheck that c g is a generalized capacity on T. Hence we are reduced to provingthe theorem when Ω is a compact space and E is a Fσδ subset of Ω. Then write

E =⋂

ℓ>1

Gℓ, Gℓ =⋃

m>1

Kℓm

where Kℓm is a closed subset of Ω. Without loss of generality, we can arrange thatKℓm is increasing in m. Fix λ < c(E). Then

E = G1 ∩⋂

ℓ>2

Gℓ =⋃

m>1

(K1m ∩⋂

ℓ>2

Gℓ)

and axiom (b) implies that exists a subset E1 = K1m1∩

⋂ℓ>2Gℓ of E such that

c(E1) > λ. By induction, there is a decreasing sequence E ⊃ E1 ⊃ . . . ⊃ Es with

Es = K1m1∩ . . . ∩Ksms

∩⋂

ℓ>s+1

Gℓ

and c(Es) > λ. Set K =⋂Ksms

=⋂Es ⊂ E. Axiom (c) implies

c(K) = lims→+∞

c(K1ms

∩ . . . ∩Ksms

)> lims→+∞

c(Es) > λ

and the theorem is proved.

Now, we apply these general results to the outer Monge-Ampere capacity c⋆

introduced in § 6.

Theorem 9.9. — Let Ω ⊂⊂ Cn be strongly pseudoconvex. Then the outercapacity c⋆(•,Ω) is a generalized capacity in the sense of 9.7.

Proof. — Axiom 9.7 (a) is clear, and 9.7 (c) is a consequence of 8.3. To prove9.7 (b), we only have to show that c⋆(

⋃Ej ,Ω) 6 limj→+∞ c⋆(Ej,Ω). It is no loss of

generality to assume that Ej ⊂⊂ Ω. Let Nj be the negligible set Nj = uEj< u⋆Ej

and G0 an open subset of Ω with G0 ⊃⋃Nj and c(G0,Ω) < ε. Consider the open

sets Vj = u⋆Ej< −1 + η and Gj = G0 ∪ Vj ⊃ Ej . Then (1 − η)−1u⋆Ej

6 u⋆Vj6 0

and lemma 8.4 implies

c(Gj ,Ω) 6 ε+ c(Vj ,Ω) = ε+

∫

Ω

(ddcu⋆Vj)n

6 ε+ (1 − η)−n∫

Ω

(ddcu⋆Ej)n = ε+ (1 − η)−nc⋆(Ej,Ω),

thanks to proposition 8.5. Further Ej ⊂ Gj and G1 ⊂ G2 ⊂ . . . since u⋆Ejis

decreasing. Thus G =⋃Gj ⊃ E =

⋃Ej and

c(G,Ω) = limj→+∞

c(Gj ,Ω) 6 ε+ (1 − η)−n limj→+∞

c⋆(Ej,Ω).

Letting ε, η → 0 we get the desired inequality

c⋆(E,Ω) 6 limj→+∞

c⋆(Ej,Ω).

Choquet’s capacitability theorem combined with 8.3 (c) implies that every K-analytic subset E ⊂ Ω satisfies

(9.10) c⋆(E,Ω) = supK compact⊂E

c(K,Ω) = c⋆(E,Ω).

When E is a Borel set, we thus have c⋆(E,Ω) = c(E,Ω). These results can berestated as :

Theorem 9.11. — Every K-analytic (in particular every Borel) subset of Ωis c-capacitable. Consequently c is outer regular.

10. Siciak’s extremal function and Alexander’s capacity.

We work here on the whole space Cn rather than on a bounded open subset Ω.

In this case, the relevant class of psh functions to consider is the set Plog(Cn) of

psh functions v with logarithmic growth at infinity, i.e. such that

(10.1) v(z) 6 log+ |z| + C

for some real constant C. Let E be a bounded subset of Cn. We consider the globalextremal function introduced by Siciak [Sic] :

(10.2) UE(z) = supv(z) ; v ∈ Plog(Cn), v 6 0 on E.

Theorem 10.3. — If U⋆E is not identically +∞, then U⋆E ∈ Plog(Cn) and U⋆E

satisfies an inequality

log+(|z|/R) 6 U⋆E(z) 6 log+ |z| + C

for suitable constants C,R > 0. Moreover U⋆E = 0 on E0,

(ddcU⋆E) = 0 on Cn \ E,

∫

E

(ddcU⋆E)n = 1.

Proof. — Assume that U⋆E is not identically +∞. Then U⋆E(z0) < +∞ for somepoint z0 ∈ C

n, say z0 = 0 for simplicity. By the upper semi-continuity of U⋆E , thisimplies that all functions v ∈ Plog(C

n) such that v 6 0 on E are uniformly boundedabove by some constant M0 on a small ball B(0, r0). As supB(0,r) v = χ(log r)where χ is a convex increasing function, (10.1) implies χ′ 6 1, thus

χ(log r) 6 χ(log r0) + log r/r0 for r > r0.

Therefore all functions v under consideration satisfy v(z) 6 log+(|z|/r0) + M0.In particular these functions are uniformly bounded above everywhere and U⋆E =(supv)⋆ is psh of logarithmic growth on C

n. On the other hand, E is containedin a ball B(0, R) so log+ |z|/R 6 0 on E and we get UE(z) > log+ |z|/R. The

equality (ddcU⋆E)n = 0 on Cn \ E is verified exactly in the same way as 7.8 (a).The integral over E is obtained by the following lemma.

Lemma 10.4. — Let v ∈ Plog(Cn) be such that

log+ |z| − C1 6 v(z) 6 log+ |z| + C2

for some constants. Then∫

Cn(ddcv)n = 1.

Proof. — It is sufficient to check that∫

Cn

(ddcv1)n 6

∫

Cn

(ddcv2)n

when v1, v2 are two such functions. Indeed, we have∫

Cn

(ddc log+ |z|)n =

∫

Cn

(ddc log |z|)n = 1

by Stokes’ theorem and remark 2.10, and we only have to choose v1(z) orv2(z) = log+ |z| and the other function equal to v. To prove the inequality, fixr, ε > 0 and choose C > 0 large enough so that (1 − ε)v1 > v2 − C on B(0, r). Asthe function u = max(1 − ε)v1, v2 − C is equal to v2 − C for |z| = R large, weget

(1 − ε)n∫

B(0,r)

(ddcv1)n =

∫

B(0,r)

(ddcu)n 6

∫

B(0,R)

(ddcu)n =

∫

B(0,R)

(ddcv2)n

and the expected inequality follows as ε→ 0 and r → +∞.

Theorem 10.5. — Let E,E1, E2, . . . ⊂ B(0, R) ⊂ Cn.

(a) If E1 ⊂ E2, then U⋆E1> U⋆E2

.

(b) If E1 ⊂ E2 ⊂ . . . and E =⋃Ej , then U⋆E = lim ↓ U⋆Ej

.

(c) If K1 ⊃ K2 ⊃ . . . and K =⋂Kj, then U⋆K = (lim ↑ U⋆Kj

)⋆.

(d) For every set E, there exists a decreasing sequence of open sets Gj ⊃ E suchthat U⋆E = (lim ↑ U⋆Gj

)⋆.

Proof. — (a) is obvious and the proof of (c) is similar to that of 8.3 (a).

(d) By Choquet’s lemma, there is an increasing sequence vj ∈ Plog(Cn) with

U⋆E = (lim vj)⋆ and vj(z) > log+ |z|/R. Set Gj = vj < 1/j and observe that

U⋆Gj> vj − 1/j.

(b) Set v = lim ↓ U⋆Ej. Then v ∈ Plog(C

n) and v = 0 on E, except on the

negligible set N =⋃UEj

< U⋆Ej. By Josefson’s theorem, there exists w ∈ P (Cn)

such that N ⊂ w = −∞. We set

vj(z) =

(1 − 1

j

)v(z) + maxεjw(z), 1

j log |z| − j on B(0, ej3

)(1 − 1

j

)v(z) + 1

jlog |z| − j on Cn \B(0, ej

3

)

where εj is chosen such that εjw < j2 − j on S(0, ej3

). Then vj ∈ Plog(Cn) and

vj 6 0 everywhere on E for j large. Therefore

U⋆E > UE > vj >(1 −

1

j

)v + εjw on B(0, ej

3

)

and letting j → +∞ we obtain U⋆E > v. The other inequality is clear.

Now, we show that the extremal function of a compact set can be computed interms of polynomials. We denote by Pd the space of polynomials of degree 6 d inC[z1, . . . , zn].

Theorem 10.6. — Let K be a compact subset of Cn. Then

UK(z) = sup1

dlog |P (z)| ; d > 1, P ∈ Pd, ‖P‖L∞(K) 6 1

.

Proof. — For any of the polynomials P involved in the above formula, weclearly have 1

dlog |P | ∈ Plog(C

n) and this function is 6 0 on K. Hence

1

dlog |P | 6 UK .

Conversely, fix a point z0 ∈ Cn and a real number a < UK(z0). Then there exists

v ∈ Plog(Cn) such that v 6 0 on K and v(z0) > a. Replacing v by v ⋆ ρδ − ε

with δ ≪ ε ≪ 1, we may assume that v ∈ Plog(Cn) ∩ C∞(Cn), v < 0 on K

and v(z0) > a. Choose a ball B(z0, r) on which v > a, a smooth function χ withcompact support in B(z0, r) such that χ = 1 on B(z0, r/2) and apply Hormander’sL2 estimates to the closed (0, 1)-form d′′χ and to the weight

ϕ(z) = 2dv(z) + 2n log |z − z0| + ε log(1 + |z|2).

We find a solution f of d′′f = d′′χ such that∫

Cn

|f |2e−2dv|z − z0|−2n(1 + |z|2)−εdλ

6

∫

B(z0,r)

|d′′χ|2e−2dv|z − z0|−2n(1 + |z|2)2−εdλ 6 C1e

−2da.

We thus have f(z0) = 0 and F = χ− f is a holomorphic function on Cn such that

F (z0) = 1. In addition we get∫

Cn

|F |2e−2dv(1 + |z|2)−2n−2εdλ 6 C2e−2da

where C1, C2 > 0 are constants independent of d. As v(z) 6 log+ |z| + C3, itfollows that F ∈ Pd. Moreover, since v > 0 on a neighborhood of K, the meanvalue inequality applied to the subharmonic function |F |2 gives

supK

|F |2 6 C4e−2da.

The polynomial P = C−1/24 edaF ∈ Pd is such that ‖P‖K = 1 and we have

log |P (z0)| > da− C5, whence

sup1

dlog |P (z0)| ; d > 1, P ∈ Pd, ‖P‖L∞(K) 6 1

> a.

As a was an arbitrary number < UK(z0), the proof is complete.

Now, we introduce a few concepts related to extremal polynomials. Let B bethe unit ball of Cn and K a compact subset of B. The Chebishev constants Md(K)are defined by

(10.7) Md(K) = inf‖P‖L∞(K) ; P ∈ Pd, ‖P‖L∞(B) = 1

.

It is clear that Md(K) 6 1 and that Md(K) satisfies

Md+d′(K) 6 Md(K)Md′(K).

The Alexander capacity is defined by

(10.8) T (K) = infd>1

Md(K)1/d.

It is easy to see that we have in fact T (K) = limd→+∞Md(K)1/d : for any integerδ > 1, write δ = qd+ r with 0 6 r < d and observe that

Mδ(K)1/δ 6 Mqd(K)1/(qd+r) 6 Md(K)q/(qd+r) ;

letting δ → +∞ with d fixed, we get

T (K) 6 lim infδ→+∞

Mδ(K)1/δ 6 lim supδ→+∞

Mδ(K)1/δ 6 Md(K)1/d,

whence the equality. Now, for an arbitrary subset E ⊂ B, we set

(10.9) T⋆(E) = supK⊂E

T (K) , T ⋆(E) = infG open⊃E

T⋆(G).

Siciak’s theorem 10.10. — For every set E ⊂ B,

T ⋆(E) = exp(− supBU⋆E).

Proof. — The main step is to show that the equality holds for compact subsetsK ⊂ B, i.e. that

(10.11) T (K) = exp(− supBU⋆K).

Indeed, it is clear that supB U⋆K = supB UK and theorem 10.6 gives

supBUK = sup

1

dlog ‖P‖L∞(B) ; d > 1, P ∈ Pd, ‖P‖L∞(K) = 1

= sup−

1

dlog ‖P‖L∞(K) ; d > 1, P ∈ Pd, ‖P‖L∞(B) = 1

after an obvious rescaling argument P 7→ αP. Taking the exponential, we get

exp(− supBU⋆K) = inf

d>1inf

‖P‖

1/dL∞(K) ; P ∈ Pd, ‖P‖L∞(B) = 1

= infd>1

Md(K)1/d = T (K).

Next, let G be an open subset of B and Kj an increasing sequence of compactsets such that G =

⋃Kj and T⋆(G) = limT (Kj). Then 10.5 (b) implies

U⋆G = lim ↓ U⋆Kj, hence

limj→+∞

supBU⋆Kj

= supBU⋆G = sup

B

U⋆G

by Dini’s lemma. Taking the limit in (10.11), we get

T⋆(G) = exp(− supBU⋆G).

Finally, 10.5 (d) shows that there exists a decreasing sequence of open sets Gj ⊃ Esuch that U⋆E = (lim ↑ U⋆Gj

)⋆. We may take Gj so small that T ⋆(E) = limT⋆(Gj).Theorem 10.9 follows.

Corollary 10.12. — The set function T ⋆ is a generalized capacity in thesense of definition 9.7 and we have T ⋆(E) = T⋆(E) for every K-analytic set E ⊂ B.

Proof. — Axioms 9.7 (a,b,c) are immediate consequences of properties 10.5(a,b,c) respectively. In addition, formulas 10.10 and 10.11 show that T ⋆(K) =T (K) for every compact set K ⊂ B. The last statement is then a consequence ofChoquet’s capacitability theorem.

To conclude this section, we show that 1/| logT ⋆| is not very far from beingsubadditive. We need a lemma.

Lemma 10.13. — For every P ∈ Pd, one has

log ‖P‖L∞(B) − cnd 6

∫

∂B

log |P (z)|dσ(z) 6 log ‖P‖L∞(B)

where dσ is the unit invariant measure on the sphere and cn a constant such thatcn ∼ log(2n) as n→ +∞.

Proof. — Without loss of generality, we may assume that ‖P‖L∞(B) = 1. Since1d log |P | ∈ Plog(C

n), the logarithmic convexity property already used implies that

supB(0,r)

1

dlog |P | > log r for r < 1.

The Harnack inequality for the Poisson kernel implies now

supB(0,r)

log |P | 61 − r2

(1 + r)2n

∫

∂B

log |P |dσ,

∫

∂B

log |P |dσ >1 + r2n

1 − r2log r.d.

The lemma follows with

cn = infr∈]0,1[

(1 + r)2n

1 − r2log

1

r;

the infimum is attained approximately for r = 1/(2n log 2n).

Corollary 10.14. — For Pj ∈ Pdj, 1 6 j 6 N,

‖P1 . . . PN‖L∞(B) > e−cn(d1+···+dN )‖P1‖L∞(B) . . . ‖PN‖L∞(B).

Proof. — Apply lemma 10.13 to each Pj and observe that∫

∂B

log |P1 . . . PN |dσ =∑

16j6N

∫

∂B

log |Pj |dσ.

Theorem 10.15. — For any set E =⋃j>1 Ej, one has

1

cn − logT ⋆(E)6

∑

j>1

1

| logT ⋆(Ej)|.

Proof. — It is sufficient to check the inequality for a finite union K =⋃Kj

of compacts sets Kj ⊂ B, 1 6 j 6 N. Select Pj ∈ Pdjsuch that

‖Pj‖L∞(B) = 1, ‖Pj‖L∞(Kj) = Mdj(Kj),

and set P = P1 . . . PN , d = d1 + · · · + dN . Then corollary 10.14 shows that‖Pj‖L∞(B) > e−cnd, thus

Md(K) 6 ecnd‖P‖L∞(K).

If z ∈ K is in Kj , then |P (z)| 6 |Pj(z)| 6 ‖Pj‖L∞(Kj) because all other factorsare 6 1. Thus

Md(K) 6 ecnd max‖Pj‖L∞(Kj),

T (K) 6 Md(K)1/d 6 ecn maxMdj(K)1/dj ·dj/d.

Take dj = [kαj ] with arbitrary αj > 0 and let k → +∞. It follows that

T (K) 6 ecn maxT (Kj)αj/α

where α =∑αj. The inequality asserted in theorem 10.15 is obtained for the

special choice αj = 1/| logT (Kj)| > 0 which makes all terms in max. . . equal.

11. Comparison of capacities and El Mir’s theorem.

We first prove a comparison theorem for the capacities c(•,Ω) and T, due toAlexander and Taylor [A–T].

Theorem 11.1. — Let K be a compact subset of the unit ball B ⊂ Cn. Then

(a) T (K) 6 exp(−c(K,B)−1/n).

For each r < 1, there is a constant A(r) such that

(b) T (K) > exp(−A(r) c(K,B)−1) when K ⊂ B(0, r).

Remark 11.2. — Both the set functions c⋆(•,Ω) and T ⋆ are generalized ca-pacities. Hence, the estimates of the theorem also hold for all K-analytic sets, inparticular all Borel sets.

Remark 11.3. — The inequalities are sharp, at least as far as the exponents onc(K,B) are concerned. For if K = B(0, ε), then it is easy to check that T (K) = εand exercise 7.13 gives c(K,B) = (log 1/ε)−n. Hence, equality holds in (11.2). Onthe other hand, if K is a small polydisc

K = (z1, . . . , zn) ∈ Cn ; |z1| 6 δ, |zj | 6 1/n, 1 < j 6 n

and δ 6 1/n, then T (K) 6 δ, while c(K,B) > C(log 1/δ)−1. To check the lastinequality, put

u(z) =(

log1

δ

)−1

log+

|z1|

δ+ (logn)−1

n∑

j=2

log+(n|zj |).

Then u = 0 and u 6 n on B, hence

c(K,B) >

∫

B

( 1

nddcu

)n=

n!

nn(logn)−(n−1)(log 1/δ)−1

because all measures ddc log+(|zj |/r) have total mass 1 in C.

Proof of 11.1 (a). — Set M = supB U⋆K ; then T (K) = e−M by Siciak’s

theorem. Since u = U⋆K/M ∈ P (B) and 0 6 u 6 1 on B, we get

c(K,B) > M−n

∫

K

(ddcU⋆K)n = M−n

by theorem 10.3. This inequality is equivalent to 11.1 (a).

Proof of 11.1 (b). — Let u⋆K be the extremal function for K relative to theball B′ = B(0, e) ⊃⊃ B. For any v ∈ Plog(C

n) such that v 6 0 on K, we havev 6 U⋆K 6 M + 1 on B′, hence the function

w =v −M − 1

M + 1

satisfies w 6 0 on B′ and w 6 −1 on K. We infer w 6 u⋆K ; by taking thesupremum over all choices of v, we get

u⋆K >U⋆K −M − 1

M + 1.

Now, there is a point z0 ∈ B such that U⋆K(z0) = M, thus

u⋆K(z0) > −1

M + 1.

As u⋆K 6 0 on B′, the mean value and Harnack inequalities show that

u⋆K(z0) 6 C1

∫

B′

u⋆Kdλ =⇒ ‖u⋆K‖L1(B′) 6 −1

C1u⋆K(z0) 6

C2

M.

The Chern-Levine-Nirenberg inequalities 1.3 and 1.4 (a) imply now

c(K,B′) =

∫

B

(ddcu⋆K)n 6 C3‖u⋆K‖L1(B′) ‖u

⋆K‖n−1

L∞(B′) 6C4

M.

As K ⊂ B(0, r) ⊂⊂ B, theorem 6.5 (b) gives

c(K,B) 6 C5(r)c(K,B′) 6 A(r)M−1

and inequality 1.11 (b) follows.

We now prove El Mir’s theorem [E–M]. This result is an effective version ofJosefson’s theorem : given a psh function in the ball, a subextension can be foundwith prescribed singularities of poles and slow growth at infinity.

El Mir’s theorem 11.4. — Let v ∈ P (B) with v 6 −1, ε ∈ ]0, 1/n[ and

r < 1. Then there exists u ∈ Plog(Cn) such that u 6 −|v|

1n−ε on B(0, r).

Proof. — For t > 1, set Gt = z ∈ B(0, r); v(z) < −t and let U⋆t ∈ Plog(Cn)

be the Siciak extremal function of Gt. Since Gt is open, we have U⋆t = 0 on Gt.We set M(t) = supB U

⋆t and

u(z) = ε−1

∫ +∞

1

t−1−ε(U⋆t (z) −M(t))dt.

Theorem 6.6 shows that c(Gt, B) 6 C1/t, therefore

M(t) = − log T ⋆(Gt) > c(Gt, B)−1/n > C2t1/n

by inequality 1.11 (a). As U⋆t −M(t) 6 0 on B, we get U⋆t (z) −M(t) 6 log+ |z|by logarithmic convexity, thus

u(z) 6 log+ |z|.

For z ∈ B(0, r) we have U⋆t (z) = 0 as soon as Gt ∋ z, i.e. t < −v(z). Hence

u(z) 6 −ε−1

∫ |v(z)|

1

t−1−εM(t)dt 6 −C3

∫ |v(z)|

1

t−1−ε+ 1n dt = −C4|v(z)|

1n−ε.

Starting if necessary with a smaller value of ε and subtracting a constant to u, wecan actually get

u 6 −|v|1n−ε on B(0, r).

It remains to check that u is not identically −∞. By logarithmic convexity again,we have

supB(0,1/2)

U⋆t > M(t) − log 2

and there exists z0 ∈ S(0; 1/2) such that U⋆t (z0) −M(t) > − log 2. The Harnackinequality shows that

1 − 1/4

(1 + 1/2)2n

∫

∂B

(U⋆t (z) −M(t)

)dσ(z) > U⋆t (z0) −M(t) > − log 2

and integration with respect to t yields∫

∂B

u(z)dσ(z) > −4/3(3/2)2n log 2 > −∞.

References

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[B-T1] E. Bedford and B.A. Taylor. — The Dirichlet problem for a complex Monge-Ampere equation, Invent. Math., 37 (), 1–44.

[B-T2] E. Bedford and B.A. Taylor. — A new capacity for plurisubharmonic func-tions, Acta Math., 149 (), 1–41.

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Jean-Pierre DemaillyUniversite de Grenoble I,Institut Fourier, BP 74,Laboratoire associe au C.N.R.S. n 188,F-38402 Saint-Martin d’Heres